Question

A hollow ball with radius R = 2 cm has a charge of -2 nC spread uniformly over its surface (see the figure). The center of the ball is at P1 = <-1, 0, 0> cm. A point charge of 6 nC is located at P3 = <7, 0, 0> cm.

What is the net electric field at location P2 = <0, 4, 0> cm?

At a particular instant an electron is at location P2. What is the net electric force on the electron at that instant? .

Answer 1

Given that,

R = 2 cm

q1 = -2 nC

for partice 1,

x1 = -1 cm , y1 = 0 , z1 = 0

q2 = 6 nC

for particle 2,

x2 = 7 cm , y2 = 0 , z2 = 0

x = 0 , y = 4 cm , z = 0

Electric field due to particle 1,

E1 = kq₁ / [(x - x₁)² + (y - y₁)²)]^3/2 * [(x - x₁) i + (y - y₁) j]

E1 = 9*10^9*2*10^(-9) / [(0.01)² + (0.04)²]^3/2 * [0.01i + 0.04j]

E1 = 2568.02i + 10272.0j

Electric field due to particle 2,

E2 = kq2/ [(x - x2)² + (y - y2)²)]^3/2 * [(x - x2) i + (y - y2) j]

E2 = 9*10^9*6*10^(-9) / [(0 - 0.07)² + (0.04)²]^3/2 * [-0.07i + 0.04j]

E2 = -7213.09i + 4121.76j

Net electric field,

E = E1 + E2 = -4645.07i + 14393.84

net electric field at location P2 = <0, 4, 0>,

E = sqrt (Ex^2 + Ey^2)

E = 15.12*10^3 N/C

(b)

F = qE

F = 1.6*10^(-19) * 15.12*10^3

F = 24.19*10^(-16) N