Answer:
-54.12 V
Explanation:
The work done by this force is equal to the difference between the final value and the initial value of the energy. Since the charge starts from the rest its initial kinetic energy is zero.
[tex]W=\Delta E\\W=\Delta K+\Delta U\\W=K_f+\Delta U\\\Delta U=W-K_f\\\Delta U=6.10*10^{-4}J-1.50*10^{-4}J\\\Delta U=4.60*10^{-4}J[/tex]
The change in electrostatic potential energy [tex]\Delta U[/tex], of one point charge q is defined as the product of the charge and the potential difference.
[tex]\Delta U=qV\\V=\frac{\Delta U}{q}\\V=\frac{4.60*10^{-4}J}{-8.50*10^{-6}C}\\V=-54.12 V[/tex]
The potential difference between a and b will be "-54.12 V".
According to the question,
Charge = -8.50 μC
Work done = 6.10×10⁻⁴ J
As we know,
→ W = ΔE
or,
= ΔK + ΔU
[tex]= K_f[/tex] + ΔU
Now,
→ ΔU = W - [tex]K_f[/tex]
By substituting the values, we get
[tex]= 6.10\times 10^{-4} -1.50\times 10^{-4}[/tex]
[tex]= 4.60\times 10^{-4} \ J[/tex]
hence,
The potential difference be:
→ [tex]\Delta U = qV[/tex]
[tex]V = \frac{\Delta U}{q}[/tex]
[tex]= \frac{4.60\times 10^{-4}}{-8.50\times 10^{-6}}[/tex]
[tex]= -54.12 \ V[/tex]
Thus the above response is right.
Find out more information about Kinetic energy here:
https://brainly.com/question/25959744
