The potentials of the two conductors are 4V and -8V. What work is done by the electric field when the charge

The potentials of the two conductors are 4V and -8V. What work is done by the electric field when the charge moves + 2.5 * 10 ^ -7 C from the first conductor to the second?

φ2 = -8 V.

q = + 2.5 * 10 ^ -7 Cl.

A -?

The work of the electric field A on the movement of the charge is determined by the formula: A = q * (φ1 – φ2), where q is the magnitude of the electric charge, φ1, φ2 are the potential of the electric field at the initial and final point of movement.

Since the charge is positive, and it is moved from negative potential to positive, it is the electric field that does the work.

A = 2.5 * 10 ^ -7 C * (4 V – (-8 V) = 80 * 10 ^ -7 J.

Answer: when the charge moves, the electric field does work A = 80 * 10 ^ -7 J.