1. Two point charges are located on the x-axis as shown, a +2 µC charge at x = -5 cm, and a - 2µC charge at x =+5 cm. (a) Calculate the electric potential at the origin, x= y = 0 cm. (b) For what values of x and y would the electric potential be the same as at the origin? (c) Based upon this equipotential line, what direction should the Electric field be in along this line? (d) If another point charge is placed at y =+5 cm, find the electric potential energy of this particle if it has a total charge of +2 nC.

Principles of Physics: A Calculus-Based Text
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Chapter20: Electric Potential And Capacitance
Section: Chapter Questions
Problem 17P: Two particles each with charge +2.00 C are located on the x axis. One is at x = 1.00 m, and the...
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Please answer question (d)
1. Two point charges are located on the x-axis as shown, a +2 µC charge at x = -5 cm, and a -
2µC charge at x=+5 cm.
(a) Calculate the electric potential at the origin, x=y = 0 cm.
(b) For what values of x and y would the electric potential be the same as at the origin?
(c) Based upon this equipotential line, what direction should the Electric field be in along this line?
(d) If another point charge is placed at y =+5 cm, find the electric potential energy of this particle if it has
a total charge of +2 nC.
Transcribed Image Text:1. Two point charges are located on the x-axis as shown, a +2 µC charge at x = -5 cm, and a - 2µC charge at x=+5 cm. (a) Calculate the electric potential at the origin, x=y = 0 cm. (b) For what values of x and y would the electric potential be the same as at the origin? (c) Based upon this equipotential line, what direction should the Electric field be in along this line? (d) If another point charge is placed at y =+5 cm, find the electric potential energy of this particle if it has a total charge of +2 nC.
Expert Solution
Step 1
Electric potential = 14TE0xQr
Step 2
(a)
Electric potential at origin due to two charges
V0,0=14TTE02×10-65×10-2+2×10-65×10-2=14Ttɛ02x2x10-65x10-2=9×1094×10-65x10-
2=7.2×105V
(b)
Let V(x,y)=V(0,0).
14TE02×10-6x+0.052+y2+2×10-6x-0.052+y2=14T1E02×2×10-65×10-21x+0.052+y2+1x-
0.052+y2=25x10-2
The shape of this equipotential line is plotted below
2.
10°
0,04
0.02
-O DB
-0 06
-0 04
-0 02
0 02
0 04
0 06
0.02
-0 04
(c) The electric field will be perpendicular to the red line
Step 3
According to Bartleby guidelines only three paerts of a question can be answered at once.
Please post the remaining questions separately.
Rate this solution
Transcribed Image Text:Expert Solution Step 1 Electric potential = 14TE0xQr Step 2 (a) Electric potential at origin due to two charges V0,0=14TTE02×10-65×10-2+2×10-65×10-2=14Ttɛ02x2x10-65x10-2=9×1094×10-65x10- 2=7.2×105V (b) Let V(x,y)=V(0,0). 14TE02×10-6x+0.052+y2+2×10-6x-0.052+y2=14T1E02×2×10-65×10-21x+0.052+y2+1x- 0.052+y2=25x10-2 The shape of this equipotential line is plotted below 2. 10° 0,04 0.02 -O DB -0 06 -0 04 -0 02 0 02 0 04 0 06 0.02 -0 04 (c) The electric field will be perpendicular to the red line Step 3 According to Bartleby guidelines only three paerts of a question can be answered at once. Please post the remaining questions separately. Rate this solution
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