A small sphere contains a charge of q_1 = 5.0 mu C as shown below. What will be

Question

A small sphere contains a charge of q_1 = 5.0 mu C as shown below. What will be the direction and magnitude of the electric field at point A? E = 9 times 10^9 times (5 times 10^-6)/0.03^2 = 5 times 10^7 (N/C), 0 degree from What will be the electrostatic potential at point A? V_A = 9 times 10^9 times (5 times 10^-6)/0.03 = 1.5 times 10^6 (J/C) What will be the force felt by a proton placed at B? F = 9 times 10^9 times (5 times 10^-6) (1.6 times 10^-19)/(0.03 + 0.09)^2 = 5 times 10^-13 N How much work would be required to move a proton from point B to point A? V_B = 9 times 10^9 times (5 times 10^-6)/0.12 = 3.75 times 10^5 (J/C) W_BA = 5 times 10^-6 (3.75 times 10^5 - 1.5 times 10^6) = -5.6 (J)

Explanation / Answer

Ans Force experienced by proton at point A=Fa= k*q*Q/r^2

Fa=9*10^9*5*10^-6*1.6*10^-19/(3*10^-2)^2

Fa= 8*10^-12 N

Fb=5*10^-13 N (already calculated)

net force experienced by proton =Fa-Fb

Fnet=(8*10^-12)-(5*10^-13)

Fnet=7.5*10^-12

work= Fnet*displacement=7.5*10^-12*9*10^-2

work=6.75*10^-13 N

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