Question

The work done by an external force to move a -6.70 μc charge from point a to point b is 1.20×10−3 j .

Answer 1

Answer:

108.66V

Explanation:

E = 1.20*10⁻³J

q = 6.70μC = 6.7*10⁻⁶C

V =?

K.E = 4.72 * 10⁻⁴J

E = q∇v

but there's difference in energy in moving the electron from point A to B. The electron has an initial energy of 4.72*10⁻⁴J.

E = Q - KE

E = 1.20*10⁻³ - 4.72*10⁻⁴ = 7.28*10⁻⁴J

E = q∇v

∇v = E / q

∇v = 7.28*10⁻⁴ / 6.7*10⁻⁶

∇v = 108.66V

the change in potential difference from point A to B was 108.66V

Answer 2

Answer:

108.7 V

Explanation:

Two forces are acting on the particle:

- The external force, whose work is $W=1.20 \cdot 10^{-3}J$

- The force of the electric field, whose work is equal to the change in electric potential energy of the charge: $W_e=q\Delta V$

where

q is the charge

$\Delta V$ is the potential difference

The variation of kinetic energy of the charge is equal to the sum of the work done by the two forces:

$K_f - K_i = W + W_e = W+q\Delta V$

and since the charge starts from rest, $K_i = 0$, so the formula becomes

$K_f = W+q\Delta V$

In this problem, we have

$W=1.20 \cdot 10^{-3}J$ is the work done by the external force

$q=-6.70 \mu C=-6.7\cdot 10^{-6}C$ is the charge

$K_f = 4.72\cdot 10^{-4}J$ is the final kinetic energy

Solving the formula for $\Delta V$, we find

$\Delta V=\frac{K_f-W}{q}=\frac{4.72\cdot 10^{-4}J-1.2\cdot 10^{-3} J}{-6.7\cdot 10^{-6}C}=108.7 V$