Two point charges both -1 mu C are separated by 2 cm as shown in box A. What is

Question

Two point charges both -1 mu C are separated by 2 cm as shown in box A. What is the resultant electric potential V at the point P that lies 3 cm below the midpoint of the two charges in box A (taking V = 0 at infinity)? If the two charges are replaced with a 2-cm-long uniformly charged line with the same total charge (-2 mu C), with midpoint 3 cm above P (as shown in box B), will the magnitude of the electric field at P be larger, smaller, or the same as it was in box A? Explain your reasoning. What is the electric field at point P in box A (express your answer in vector notation) ? An electron is moving with velocity 5 cm/s in the +x direction in an electric field E = 3 V/cm that points In the -y direction. What is the direction and magnitude of the electron's acceleration?

Explanation / Answer

problem 1:

Electric potential Due to each point charge is V = Kq/r

Here due to two point charges

Vnet = V1 + V2

as V1 = V2

Vnet = 2V

since q1 = q2 = -1uC

Distance of Point P from each charge is

r^2 = 3^2 + (2/2)^2

r = 3.16 cm or 0.0316 m

so

Vnet = 2* 9*10^9 * -1*10^-6/(0.0316)

Vnet = 5.7*10^5 Volts

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part B:

This Follows from Uniform charge Distirbution

thst EF due to Cylindrical Symmetry is E = L/2pieo R

Where L is Charge per unit length

thus

E increases are net Charge Ditribution is Enhanced.

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part A :

Electric field due to each point charge is E = Kq/r^2

Ex Due to one charge is Ex = 9*10^9 * -1*10^-6/(0.01^2)

Ex = -9 *10^7 N/C

Ex net due to both Charges = E1 - E2 = 0

as they are like charges, Enet due to x comp is 0.

point P is at distance of 0.0316 m from each charge

so

Enet Due to both the charges is Ey = 2 KQ/r^2

Ey = 2 * 9*10^9 * -1*10^-6/0.0316^2

Ey = -1.802 *10^7 N/C

so

Enet = 0i - 1.8 *10^7 j   

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