Question

A uniform electric field, with a magnitude of 370 N/C, is directed parallel to the positive x-axis. If the electric potential at x = 2.00 m is 1 000 V, what is the change in potential energy of a particle with a charge of + 2.80 x 10-3 C as it moves from x = 1.9 m to x = 2.1 m?

Answer 1

Answer:

$\Delta U = 0.2072 J$

Explanation:

Potential difference between two points in constant electric field is given by the formula

$\Delta V = E.\Delta x$

here we know that

$E = 370 N/C$

also we know that

$\Delta x = 2.1 - 1.9 = 0.2 m$

now we have

$\Delta V = 370 (0.2) = 74 V$

now change in potential energy is given as

$\Delta U = Q\Delta V$

$\Delta U = (2.80 \times 10^{-3})(74)$

$\Delta U = 0.2072 J$