Answer:
108.7 V
Explanation:
Two forces are acting on the particle:
- The external force, whose work is [tex]W=1.20 \cdot 10^{-3}J[/tex]
- The force of the electric field, whose work is equal to the change in electric potential energy of the charge: [tex]W_e=q\Delta V[/tex]
where
q is the charge
[tex]\Delta V[/tex] 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:
[tex]K_f - K_i = W + W_e = W+q\Delta V[/tex]
and since the charge starts from rest, [tex]K_i = 0[/tex], so the formula becomes
[tex]K_f = W+q\Delta V[/tex]
In this problem, we have
[tex]W=1.20 \cdot 10^{-3}J[/tex] is the work done by the external force
[tex]q=-6.70 \mu C=-6.7\cdot 10^{-6}C[/tex] is the charge
[tex]K_f = 4.72\cdot 10^{-4}J[/tex] is the final kinetic energy
Solving the formula for [tex]\Delta V[/tex], we find
[tex]\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[/tex]
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
