When moving a charge in an electric field with a potential difference of 3 kV, a work of 1 J was performed.

When moving a charge in an electric field with a potential difference of 3 kV, a work of 1 J was performed. Calculate the value of this charge.

Given:
φ1 – φ2 = 3 kV = 3000 V – potential difference in an electrostatic field
A = 1 J – the work of the field to move the charge from point 1 to point 2
q -? – charge
In an electric field when the charge moves, the potential difference is found by the formula:
φ1 – φ2 = A / q
From the formula for the potential difference, we find the value of the charge:
q = A / φ1 – φ2
q = 1/3000 = 0.0003 C = 3 * 10 ^ -4 C
Answer: the amount of charge q = 3 * 10 ^ -4 C