3. Two point-like charges are placed as shown in the figure, r₁ = r₂ = 28.0 cm. Find the magnitude of the electric field at point (p) shown in the figure. Let q₁ = -13.0 µC, q2 = +42.0 N/C 60 ssf60 ss160 160 ssf60 ssf60 s sf60 ssf60 r₁ O ss ssf60 ssf60 ssf6³ 1₂ sfoc fossf60 ssf60 s sf60 ssf60 ssf60 ssf6(

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Chapter1: Units, Trigonometry. And Vectors
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3. Two point-like charges are placed as shown in the figure, r₁ = r₂
point (p) shown in the figure. Let q₁
= -13.0 μC, q2 = +42.0
N/C
FO
r₁
28.0 cm. Find the magnitude of the electric field at
160 160 s 160 ssfoto
sf60 ssf60 ssf60 ssf60 ss?
r₂
ssf60 ssf60 ssfo
sfóco² féð ssf60 ssf60
ssf60 ssf60 ssf60 ssf60
f60 ssf
Transcribed Image Text:= 3. Two point-like charges are placed as shown in the figure, r₁ = r₂ point (p) shown in the figure. Let q₁ = -13.0 μC, q2 = +42.0 N/C FO r₁ 28.0 cm. Find the magnitude of the electric field at 160 160 s 160 ssfoto sf60 ssf60 ssf60 ssf60 ss? r₂ ssf60 ssf60 ssfo sfóco² féð ssf60 ssf60 ssf60 ssf60 ssf60 ssf60 f60 ssf
Expert Solution
Step 1

Electric field due to a point charge:

The magnitude of the electric field due to a point charge at r distance from it is defined by the equation,

E=kqr2

where q is the charge and k is Coulomb's constant.

If the charge is positive, then the electric field moves away from it. If the charge is negative, then the electric field moves towards it.

 

NOTE: 

k=9×109 N m2/C21 μC=10-6 C1 cm=0.01 m

 

Given:

Charges q1=-13.0 μC=-13.0×10-6 C and q2=+42.0 μC=+42.0×10-6 C

The distance of the charges q1 and q2 from point P is r1=r2=28.0 cm=0.280 m

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