A hollow ball with radius R = 2 cm has a charge of -1 nC spread uniformly over i

Question

A hollow ball with radius R = 2 cm has a charge of -1 nC spread uniformly over its surface (see the figure). The center of the ball is at P_1 = cm. A point charge of 3 nC is located at P_3 = cm. (The diagram below is not drawn exactly to scale.) What is the net electric field at location P_2 = cm? E vector = N/C At a particular instant an electron is at location P_2. What is the net electric force on the electron at that instant? F vector = N. What is the direction of the net electric force on the electron? The force is at right angles to the direction of the electric field at P_2 There is not enough information to determine the direction of the force The force is In the same direction as the electric field at P_2 The force is opposite to the direction of the net electric field at P_2

Explanation / Answer

1) Field at P2

E = summation kq /r^2

Ex = 9e9*-1e-9/(0.01^2 +0.08^2) * 0.01/sqrt (0.01^2 +0.08^2)  + 9e9*-3e-9/(0.05^2 +0.08^2) * 0.05/sqrt (0.05^2 +0.08^2)

= -1780 N/C

Ey = 9e9*-1e-9/(0.01^2 +0.08^2) * 0.08/sqrt (0.01^2 +0.08^2)  + 9e9*3e-9/(0.05^2 +0.08^2) * 0.08/sqrt (0.05^2 +0.08^2)

= 1199 N/C

Resultant E = sqrt (1780^2 +1199^2)

= 2146 N/C answer

F = qE = 1.6e-19*2146

= 3.4336*10^-16 N answer

The force is opposite to direction of electric field.

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