Question

A proton is acted on by an uniform electric field of magnitude 183 N/C pointing in the positive y-direction. The particle is initially at rest. (a) In what direction will the charge move?
(b) Determine the work done by the electric field when the particle has moved through a distance of 4.15 cm from its initial position.
(c) Determine the change in electric potential energy of the charged particle.
(d) Determine the speed of the charged particle.

Answer 1

Answer:

Part a)

charge will move initially in +Y direction

Part b)

$W = 1.22 \times 10^{-18} J$

Part c)

$\Delta U = -1.22 \times 10^{-18} J$

Part d)

$v = 3.8 \times 10^4 m/s$

Explanation:

Part a)

Since electric field is in +Y direction so positive charge will move in the same direction

So charge will move initially in +Y direction

Part b)

Force due to electric field on the charge is given as

$F = qE$

$F = (1.6 \times 10^{-19})(183)$

$F = 2.93 \times 10^{-17} N$

now work done to move it by 4.15 cm is given as

$W = F.d$

$W = (2.93 \times 10^{-17})(0.0415)$

$W = 1.22 \times 10^{-18} J$

Part c)

As we know that work done by conservative force is negative change in potential energy

so we will have

$W = -\Delta U$

$\Delta U = -1.22 \times 10^{-18} J$

Part d)

Work done by electric field = change in kinetic energy of the charge particle

$\frac{1}{2}mv^2 = 1.22 \times 10^{-18}$

here we have

$m = 1.67 \times 10^{-27} kg$

$\frac{1}{2}(1.67 \times 10^{-27})v^2 = 1.22 \times 10^{-18}$

$v = 3.8 \times 10^4 m/s$