The electrostatic potential energy of the system of the charge.

Question

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Answer 1

Question

Chapter 23, Problem 66P

(a)

To determine

The electrostatic potential energy of the system of the charge.

(a)

Expert Solution

Answer to Problem 66P

The electrostatic potential energy of the system of the charge is 190 mJ .

Explanation of Solution

Given Data:

The charge on first particle is q1=+4.20 μC

The charge on second particle is q2=+4.20 μC

The charge on third particle is q3=+4.20 μC

Formula used:

The expression for the work required to assemble the system of charges is given as,

U=k(q1q2r 1,2+q1q3r 1,3+q2q3r 2,3)

Here, the radius of the particle is given as,

r1,2=2.50 mr2,3=2.50 mr1,3=2.50 m

Calculation:

The work required to assemble the system of charges is calculated as,

U=k( q 1 q 2 r 1,2 + q 1 q 3 r 1,3 + q 2 q 3 r 2,3 )=(8.988× 109 Nm 2/ C 2)( ( 4.20 μC )( 4.20 μC ) 2.50 m + ( 4.20 μC )( 4.20 μC ) 2.50 m + ( 4.20 μC )( 4.20 μC ) 2.50 m )=190 mJ

Conclusion:

Therefore, the electrostatic potential energy of the system of the charge is 190 mJ .

(b)

To determine

The electrostatic potential energy of the system of the charge.

(b)

Expert Solution

Answer to Problem 66P

The electrostatic potential energy of the system of the charge is −63.4 mJ .

Explanation of Solution

Given Data:

The charge on first particle is q1=+4.20 μC

The charge on second particle is q2=+4.20 μC

The charge on third particle is q3=−4.20 μC

Formula used:

The expression for the work required to assemble the system of charges is given as,

U=k(q1q2r 1,2+q1q3r 1,3+q2q3r 2,3)

Here, the radius of the particle is given as,

r1,2=2.50 mr2,3=2.50 mr1,3=2.50 m

Calculation:

The work required to assemble the system of charges is calculated as,

U=k( q 1 q 2 r 1,2 + q 1 q 3 r 1,3 + q 2 q 3 r 2,3 )=(8.988× 109 Nm 2/ C 2)( ( 4.20 μC )( 4.20 μC ) 2.50 m + ( 4.20 μC )( −4.20 μC ) 2.50 m + ( 4.20 μC )( −4.20 μC ) 2.50 m )=−63.4 mJ

Conclusion:

Therefore the electrostatic potential energy of the system of the charge is −63.4 mJ .

(c)

To determine

The electrostatic potential energy of the system of the charge.

(c)

Expert Solution

Answer to Problem 66P

The electrostatic potential energy of the system of the charge is −63.4 mJ .

Explanation of Solution

Given Data:

The charge on first particle is q1=−4.20 μC

The charge on second particle is q2=−4.20 μC

The charge on third particle is q3=+4.20 μC

Formula used:

The expression for the work required to assemble the system of charges is given as,

U=k(q1q2r 1,2+q1q3r 1,3+q2q3r 2,3)

Here, the radius of the particle is given as,

r1,2=2.50 mr2,3=2.50 mr1,3=2.50 m

Calculation:

The work required to assemble the system of charges is calculated as,

U=k( q 1 q 2 r 1,2 + q 1 q 3 r 1,3 + q 2 q 3 r 2,3 )=(8.988× 109 Nm 2/ C 2)( ( −4.20 μC )( −4.20 μC ) 2.50 m + ( −4.20 μC )( 4.20 μC ) 2.50 m + ( −4.20 μC )( 4.20 μC ) 2.50 m )=−63.4 mJ

Conclusion:

Therefore the electrostatic potential energy of the system of the charge is −63.4 mJ .

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Answer 2

Question

Chapter 23, Problem 66P

(a)

To determine

The electrostatic potential energy of the system of the charge.

(a)

Expert Solution

Answer to Problem 66P

The electrostatic potential energy of the system of the charge is 190 mJ .

Explanation of Solution

Given Data:

The charge on first particle is q1=+4.20 μC

The charge on second particle is q2=+4.20 μC

The charge on third particle is q3=+4.20 μC

Formula used:

The expression for the work required to assemble the system of charges is given as,

U=k(q1q2r 1,2+q1q3r 1,3+q2q3r 2,3)

Here, the radius of the particle is given as,

r1,2=2.50 mr2,3=2.50 mr1,3=2.50 m

Calculation:

The work required to assemble the system of charges is calculated as,

U=k( q 1 q 2 r 1,2 + q 1 q 3 r 1,3 + q 2 q 3 r 2,3 )=(8.988× 109 Nm 2/ C 2)( ( 4.20 μC )( 4.20 μC ) 2.50 m + ( 4.20 μC )( 4.20 μC ) 2.50 m + ( 4.20 μC )( 4.20 μC ) 2.50 m )=190 mJ

Conclusion:

Therefore, the electrostatic potential energy of the system of the charge is 190 mJ .

(b)

To determine

The electrostatic potential energy of the system of the charge.

(b)

Expert Solution

Answer to Problem 66P

The electrostatic potential energy of the system of the charge is −63.4 mJ .

Explanation of Solution

Given Data:

The charge on first particle is q1=+4.20 μC

The charge on second particle is q2=+4.20 μC

The charge on third particle is q3=−4.20 μC

Formula used:

The expression for the work required to assemble the system of charges is given as,

U=k(q1q2r 1,2+q1q3r 1,3+q2q3r 2,3)

Here, the radius of the particle is given as,

r1,2=2.50 mr2,3=2.50 mr1,3=2.50 m

Calculation:

The work required to assemble the system of charges is calculated as,

U=k( q 1 q 2 r 1,2 + q 1 q 3 r 1,3 + q 2 q 3 r 2,3 )=(8.988× 109 Nm 2/ C 2)( ( 4.20 μC )( 4.20 μC ) 2.50 m + ( 4.20 μC )( −4.20 μC ) 2.50 m + ( 4.20 μC )( −4.20 μC ) 2.50 m )=−63.4 mJ

Conclusion:

Therefore the electrostatic potential energy of the system of the charge is −63.4 mJ .

(c)

To determine

The electrostatic potential energy of the system of the charge.

(c)

Expert Solution

Answer to Problem 66P

The electrostatic potential energy of the system of the charge is −63.4 mJ .

Explanation of Solution

Given Data:

The charge on first particle is q1=−4.20 μC

The charge on second particle is q2=−4.20 μC

The charge on third particle is q3=+4.20 μC

Formula used:

The expression for the work required to assemble the system of charges is given as,

U=k(q1q2r 1,2+q1q3r 1,3+q2q3r 2,3)

Here, the radius of the particle is given as,

r1,2=2.50 mr2,3=2.50 mr1,3=2.50 m

Calculation:

The work required to assemble the system of charges is calculated as,

U=k( q 1 q 2 r 1,2 + q 1 q 3 r 1,3 + q 2 q 3 r 2,3 )=(8.988× 109 Nm 2/ C 2)( ( −4.20 μC )( −4.20 μC ) 2.50 m + ( −4.20 μC )( 4.20 μC ) 2.50 m + ( −4.20 μC )( 4.20 μC ) 2.50 m )=−63.4 mJ

Conclusion:

Therefore the electrostatic potential energy of the system of the charge is −63.4 mJ .

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Answer 3

Question

Chapter 23, Problem 66P

(a)

To determine

The electrostatic potential energy of the system of the charge.

(a)

Expert Solution

Answer to Problem 66P

The electrostatic potential energy of the system of the charge is 190 mJ .

Explanation of Solution

Given Data:

The charge on first particle is q1=+4.20 μC

The charge on second particle is q2=+4.20 μC

The charge on third particle is q3=+4.20 μC

Formula used:

The expression for the work required to assemble the system of charges is given as,

U=k(q1q2r 1,2+q1q3r 1,3+q2q3r 2,3)

Here, the radius of the particle is given as,

r1,2=2.50 mr2,3=2.50 mr1,3=2.50 m

Calculation:

The work required to assemble the system of charges is calculated as,

U=k( q 1 q 2 r 1,2 + q 1 q 3 r 1,3 + q 2 q 3 r 2,3 )=(8.988× 109 Nm 2/ C 2)( ( 4.20 μC )( 4.20 μC ) 2.50 m + ( 4.20 μC )( 4.20 μC ) 2.50 m + ( 4.20 μC )( 4.20 μC ) 2.50 m )=190 mJ

Conclusion:

Therefore, the electrostatic potential energy of the system of the charge is 190 mJ .

(b)

To determine

The electrostatic potential energy of the system of the charge.

(b)

Expert Solution

Answer to Problem 66P

The electrostatic potential energy of the system of the charge is −63.4 mJ .

Explanation of Solution

Given Data:

The charge on first particle is q1=+4.20 μC

The charge on second particle is q2=+4.20 μC

The charge on third particle is q3=−4.20 μC

Formula used:

The expression for the work required to assemble the system of charges is given as,

U=k(q1q2r 1,2+q1q3r 1,3+q2q3r 2,3)

Here, the radius of the particle is given as,

r1,2=2.50 mr2,3=2.50 mr1,3=2.50 m

Calculation:

The work required to assemble the system of charges is calculated as,

U=k( q 1 q 2 r 1,2 + q 1 q 3 r 1,3 + q 2 q 3 r 2,3 )=(8.988× 109 Nm 2/ C 2)( ( 4.20 μC )( 4.20 μC ) 2.50 m + ( 4.20 μC )( −4.20 μC ) 2.50 m + ( 4.20 μC )( −4.20 μC ) 2.50 m )=−63.4 mJ

Conclusion:

Therefore the electrostatic potential energy of the system of the charge is −63.4 mJ .

(c)

To determine

The electrostatic potential energy of the system of the charge.

(c)

Expert Solution

Answer to Problem 66P

The electrostatic potential energy of the system of the charge is −63.4 mJ .

Explanation of Solution

Given Data:

The charge on first particle is q1=−4.20 μC

The charge on second particle is q2=−4.20 μC

The charge on third particle is q3=+4.20 μC

Formula used:

The expression for the work required to assemble the system of charges is given as,

U=k(q1q2r 1,2+q1q3r 1,3+q2q3r 2,3)

Here, the radius of the particle is given as,

r1,2=2.50 mr2,3=2.50 mr1,3=2.50 m

Calculation:

The work required to assemble the system of charges is calculated as,

U=k( q 1 q 2 r 1,2 + q 1 q 3 r 1,3 + q 2 q 3 r 2,3 )=(8.988× 109 Nm 2/ C 2)( ( −4.20 μC )( −4.20 μC ) 2.50 m + ( −4.20 μC )( 4.20 μC ) 2.50 m + ( −4.20 μC )( 4.20 μC ) 2.50 m )=−63.4 mJ

Conclusion:

Therefore the electrostatic potential energy of the system of the charge is −63.4 mJ .

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Answer 4

Question

Chapter 23, Problem 66P

(a)

To determine

The electrostatic potential energy of the system of the charge.

(a)

Expert Solution

Answer to Problem 66P

The electrostatic potential energy of the system of the charge is 190 mJ .

Explanation of Solution

Given Data:

The charge on first particle is q1=+4.20 μC

The charge on second particle is q2=+4.20 μC

The charge on third particle is q3=+4.20 μC

Formula used:

The expression for the work required to assemble the system of charges is given as,

U=k(q1q2r 1,2+q1q3r 1,3+q2q3r 2,3)

Here, the radius of the particle is given as,

r1,2=2.50 mr2,3=2.50 mr1,3=2.50 m

Calculation:

The work required to assemble the system of charges is calculated as,

U=k( q 1 q 2 r 1,2 + q 1 q 3 r 1,3 + q 2 q 3 r 2,3 )=(8.988× 109 Nm 2/ C 2)( ( 4.20 μC )( 4.20 μC ) 2.50 m + ( 4.20 μC )( 4.20 μC ) 2.50 m + ( 4.20 μC )( 4.20 μC ) 2.50 m )=190 mJ

Conclusion:

Therefore, the electrostatic potential energy of the system of the charge is 190 mJ .

(b)

To determine

The electrostatic potential energy of the system of the charge.

(b)

Expert Solution

Answer to Problem 66P

The electrostatic potential energy of the system of the charge is −63.4 mJ .

Explanation of Solution

Given Data:

The charge on first particle is q1=+4.20 μC

The charge on second particle is q2=+4.20 μC

The charge on third particle is q3=−4.20 μC

Formula used:

The expression for the work required to assemble the system of charges is given as,

U=k(q1q2r 1,2+q1q3r 1,3+q2q3r 2,3)

Here, the radius of the particle is given as,

r1,2=2.50 mr2,3=2.50 mr1,3=2.50 m

Calculation:

The work required to assemble the system of charges is calculated as,

U=k( q 1 q 2 r 1,2 + q 1 q 3 r 1,3 + q 2 q 3 r 2,3 )=(8.988× 109 Nm 2/ C 2)( ( 4.20 μC )( 4.20 μC ) 2.50 m + ( 4.20 μC )( −4.20 μC ) 2.50 m + ( 4.20 μC )( −4.20 μC ) 2.50 m )=−63.4 mJ

Conclusion:

Therefore the electrostatic potential energy of the system of the charge is −63.4 mJ .

(c)

To determine

The electrostatic potential energy of the system of the charge.

(c)

Expert Solution

Answer to Problem 66P

The electrostatic potential energy of the system of the charge is −63.4 mJ .

Explanation of Solution

Given Data:

The charge on first particle is q1=−4.20 μC

The charge on second particle is q2=−4.20 μC

The charge on third particle is q3=+4.20 μC

Formula used:

The expression for the work required to assemble the system of charges is given as,

U=k(q1q2r 1,2+q1q3r 1,3+q2q3r 2,3)

Here, the radius of the particle is given as,

r1,2=2.50 mr2,3=2.50 mr1,3=2.50 m

Calculation:

The work required to assemble the system of charges is calculated as,

U=k( q 1 q 2 r 1,2 + q 1 q 3 r 1,3 + q 2 q 3 r 2,3 )=(8.988× 109 Nm 2/ C 2)( ( −4.20 μC )( −4.20 μC ) 2.50 m + ( −4.20 μC )( 4.20 μC ) 2.50 m + ( −4.20 μC )( 4.20 μC ) 2.50 m )=−63.4 mJ

Conclusion:

Therefore the electrostatic potential energy of the system of the charge is −63.4 mJ .

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Answer 5

Chapter 23, Problem 66P

(a)

To determine

The electrostatic potential energy of the system of the charge.

(a)

Expert Solution

Answer to Problem 66P

The electrostatic potential energy of the system of the charge is 190 mJ .

Explanation of Solution

Given Data:

The charge on first particle is q1=+4.20 μC

The charge on second particle is q2=+4.20 μC

The charge on third particle is q3=+4.20 μC

Formula used:

The expression for the work required to assemble the system of charges is given as,

U=k(q1q2r 1,2+q1q3r 1,3+q2q3r 2,3)

Here, the radius of the particle is given as,

r1,2=2.50 mr2,3=2.50 mr1,3=2.50 m

Calculation:

The work required to assemble the system of charges is calculated as,

U=k( q 1 q 2 r 1,2 + q 1 q 3 r 1,3 + q 2 q 3 r 2,3 )=(8.988× 109 Nm 2/ C 2)( ( 4.20 μC )( 4.20 μC ) 2.50 m + ( 4.20 μC )( 4.20 μC ) 2.50 m + ( 4.20 μC )( 4.20 μC ) 2.50 m )=190 mJ

Conclusion:

Therefore, the electrostatic potential energy of the system of the charge is 190 mJ .

(b)

To determine

The electrostatic potential energy of the system of the charge.

(b)

Expert Solution

Answer to Problem 66P

The electrostatic potential energy of the system of the charge is −63.4 mJ .

Explanation of Solution

Given Data:

The charge on first particle is q1=+4.20 μC

The charge on second particle is q2=+4.20 μC

The charge on third particle is q3=−4.20 μC

Formula used:

The expression for the work required to assemble the system of charges is given as,

U=k(q1q2r 1,2+q1q3r 1,3+q2q3r 2,3)

Here, the radius of the particle is given as,

r1,2=2.50 mr2,3=2.50 mr1,3=2.50 m

Calculation:

The work required to assemble the system of charges is calculated as,

U=k( q 1 q 2 r 1,2 + q 1 q 3 r 1,3 + q 2 q 3 r 2,3 )=(8.988× 109 Nm 2/ C 2)( ( 4.20 μC )( 4.20 μC ) 2.50 m + ( 4.20 μC )( −4.20 μC ) 2.50 m + ( 4.20 μC )( −4.20 μC ) 2.50 m )=−63.4 mJ

Conclusion:

Therefore the electrostatic potential energy of the system of the charge is −63.4 mJ .

(c)

To determine

The electrostatic potential energy of the system of the charge.

(c)

Expert Solution

Answer to Problem 66P

The electrostatic potential energy of the system of the charge is −63.4 mJ .

Explanation of Solution

Given Data:

The charge on first particle is q1=−4.20 μC

The charge on second particle is q2=−4.20 μC

The charge on third particle is q3=+4.20 μC

Formula used:

The expression for the work required to assemble the system of charges is given as,

U=k(q1q2r 1,2+q1q3r 1,3+q2q3r 2,3)

Here, the radius of the particle is given as,

r1,2=2.50 mr2,3=2.50 mr1,3=2.50 m

Calculation:

The work required to assemble the system of charges is calculated as,

U=k( q 1 q 2 r 1,2 + q 1 q 3 r 1,3 + q 2 q 3 r 2,3 )=(8.988× 109 Nm 2/ C 2)( ( −4.20 μC )( −4.20 μC ) 2.50 m + ( −4.20 μC )( 4.20 μC ) 2.50 m + ( −4.20 μC )( 4.20 μC ) 2.50 m )=−63.4 mJ

Conclusion:

Therefore the electrostatic potential energy of the system of the charge is −63.4 mJ .

Answer 6

Chapter 23, Problem 66P

(a)

To determine

The electrostatic potential energy of the system of the charge.

(a)

Expert Solution

Answer to Problem 66P

The electrostatic potential energy of the system of the charge is 190 mJ .

Explanation of Solution

Given Data:

The charge on first particle is q1=+4.20 μC

The charge on second particle is q2=+4.20 μC

The charge on third particle is q3=+4.20 μC

Formula used:

The expression for the work required to assemble the system of charges is given as,

U=k(q1q2r 1,2+q1q3r 1,3+q2q3r 2,3)

Here, the radius of the particle is given as,

r1,2=2.50 mr2,3=2.50 mr1,3=2.50 m

Calculation:

The work required to assemble the system of charges is calculated as,

U=k( q 1 q 2 r 1,2 + q 1 q 3 r 1,3 + q 2 q 3 r 2,3 )=(8.988× 109 Nm 2/ C 2)( ( 4.20 μC )( 4.20 μC ) 2.50 m + ( 4.20 μC )( 4.20 μC ) 2.50 m + ( 4.20 μC )( 4.20 μC ) 2.50 m )=190 mJ

Conclusion:

Therefore, the electrostatic potential energy of the system of the charge is 190 mJ .

Answer 7

Chapter 23, Problem 66P

(a)

To determine

The electrostatic potential energy of the system of the charge.

Answer 8

(a)

Expert Solution

Answer to Problem 66P

The electrostatic potential energy of the system of the charge is 190 mJ .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given Data:

The charge on first particle is q1=+4.20 μC

The charge on second particle is q2=+4.20 μC

The charge on third particle is q3=+4.20 μC

Formula used:

The expression for the work required to assemble the system of charges is given as,

U=k(q1q2r 1,2+q1q3r 1,3+q2q3r 2,3)

Here, the radius of the particle is given as,

r1,2=2.50 mr2,3=2.50 mr1,3=2.50 m

Calculation:

The work required to assemble the system of charges is calculated as,

U=k( q 1 q 2 r 1,2 + q 1 q 3 r 1,3 + q 2 q 3 r 2,3 )=(8.988× 109 Nm 2/ C 2)( ( 4.20 μC )( 4.20 μC ) 2.50 m + ( 4.20 μC )( 4.20 μC ) 2.50 m + ( 4.20 μC )( 4.20 μC ) 2.50 m )=190 mJ

Conclusion:

Therefore, the electrostatic potential energy of the system of the charge is 190 mJ .

Answer 13

(b)

To determine

The electrostatic potential energy of the system of the charge.

(b)

Expert Solution

Answer to Problem 66P

The electrostatic potential energy of the system of the charge is −63.4 mJ .

Explanation of Solution

Given Data:

The charge on first particle is q1=+4.20 μC

The charge on second particle is q2=+4.20 μC

The charge on third particle is q3=−4.20 μC

Formula used:

The expression for the work required to assemble the system of charges is given as,

U=k(q1q2r 1,2+q1q3r 1,3+q2q3r 2,3)

Here, the radius of the particle is given as,

r1,2=2.50 mr2,3=2.50 mr1,3=2.50 m

Calculation:

The work required to assemble the system of charges is calculated as,

U=k( q 1 q 2 r 1,2 + q 1 q 3 r 1,3 + q 2 q 3 r 2,3 )=(8.988× 109 Nm 2/ C 2)( ( 4.20 μC )( 4.20 μC ) 2.50 m + ( 4.20 μC )( −4.20 μC ) 2.50 m + ( 4.20 μC )( −4.20 μC ) 2.50 m )=−63.4 mJ

Conclusion:

Therefore the electrostatic potential energy of the system of the charge is −63.4 mJ .

Answer 14

(b)

To determine

The electrostatic potential energy of the system of the charge.

Answer 15

(b)

Expert Solution

Answer to Problem 66P

The electrostatic potential energy of the system of the charge is −63.4 mJ .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Given Data:

The charge on first particle is q1=+4.20 μC

The charge on second particle is q2=+4.20 μC

The charge on third particle is q3=−4.20 μC

Formula used:

The expression for the work required to assemble the system of charges is given as,

U=k(q1q2r 1,2+q1q3r 1,3+q2q3r 2,3)

Here, the radius of the particle is given as,

r1,2=2.50 mr2,3=2.50 mr1,3=2.50 m

Calculation:

The work required to assemble the system of charges is calculated as,

U=k( q 1 q 2 r 1,2 + q 1 q 3 r 1,3 + q 2 q 3 r 2,3 )=(8.988× 109 Nm 2/ C 2)( ( 4.20 μC )( 4.20 μC ) 2.50 m + ( 4.20 μC )( −4.20 μC ) 2.50 m + ( 4.20 μC )( −4.20 μC ) 2.50 m )=−63.4 mJ

Conclusion:

Therefore the electrostatic potential energy of the system of the charge is −63.4 mJ .

Answer 20

(c)

To determine

The electrostatic potential energy of the system of the charge.

(c)

Expert Solution

Answer to Problem 66P

The electrostatic potential energy of the system of the charge is −63.4 mJ .

Explanation of Solution

Given Data:

The charge on first particle is q1=−4.20 μC

The charge on second particle is q2=−4.20 μC

The charge on third particle is q3=+4.20 μC

Formula used:

The expression for the work required to assemble the system of charges is given as,

U=k(q1q2r 1,2+q1q3r 1,3+q2q3r 2,3)

Here, the radius of the particle is given as,

r1,2=2.50 mr2,3=2.50 mr1,3=2.50 m

Calculation:

The work required to assemble the system of charges is calculated as,

U=k( q 1 q 2 r 1,2 + q 1 q 3 r 1,3 + q 2 q 3 r 2,3 )=(8.988× 109 Nm 2/ C 2)( ( −4.20 μC )( −4.20 μC ) 2.50 m + ( −4.20 μC )( 4.20 μC ) 2.50 m + ( −4.20 μC )( 4.20 μC ) 2.50 m )=−63.4 mJ

Conclusion:

Therefore the electrostatic potential energy of the system of the charge is −63.4 mJ .

Answer 21

(c)

To determine

The electrostatic potential energy of the system of the charge.

Answer 22

(c)

Expert Solution

Answer to Problem 66P

The electrostatic potential energy of the system of the charge is −63.4 mJ .

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Given Data:

The charge on first particle is q1=−4.20 μC

The charge on second particle is q2=−4.20 μC

The charge on third particle is q3=+4.20 μC

Formula used:

The expression for the work required to assemble the system of charges is given as,

U=k(q1q2r 1,2+q1q3r 1,3+q2q3r 2,3)

Here, the radius of the particle is given as,

r1,2=2.50 mr2,3=2.50 mr1,3=2.50 m

Calculation:

The work required to assemble the system of charges is calculated as,

U=k( q 1 q 2 r 1,2 + q 1 q 3 r 1,3 + q 2 q 3 r 2,3 )=(8.988× 109 Nm 2/ C 2)( ( −4.20 μC )( −4.20 μC ) 2.50 m + ( −4.20 μC )( 4.20 μC ) 2.50 m + ( −4.20 μC )( 4.20 μC ) 2.50 m )=−63.4 mJ