Two charges are separated by a distance d = 5.00 cm, and Q = +7.30 nC. (a) Find

Question

Two charges are separated by a distance d = 5.00 cm, and Q = +7.30 nC.

(a) Find the electric potential at A.
1 .
Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. kV

(b) Find the electric potential at B.
2 kV

(c) Find the electric potential difference between B and A.

Explanation / Answer

1) The electric potential at a distance r from a charge q is V = q/(4*pi*epsilon*r) => the potential 2 cm from a proton is V = 1.6 * 10^(-19) /( 4*3.14*8.85*10^(-12)*2*10^(-2) ) = 0.0072 * 10^(-5) V = 7.2 * 10^(-8) V 2) The potential at 2 cm from the proton is V1 = 7.2 * 10^(-8) V. The potential at 5 cm is V2 = V1 *2/5 = 2.88 * 10^(-8) V. The potential difference is V1-V2 = 4.32 * 10^(-8) V

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