Question

Find \[\left| {{V_{BA}}} \right|\] if \[12\,J\] of work has to be done against an electric field to take a charge of \[{10^{ - 2}}\,C\] from a to b.

Answer 1

Hint:We know that in the uniform electric field, the electric potential changes with the distance from the point charge. Use the relation between electric potential, work and charge.

Formula used:
That the work required to move the charge from point a to b in the uniform electric field is,
\[W = q\Delta V\]
Here, q is the charge and \[\Delta V\] is the potential difference.

Complete step by step answer:
We know that the work required to move the charge from point a to b in the uniform electric field is,
\[W = q\Delta V\]
Here, q is the charge and \[\Delta V\] is the potential difference.
We rearrange the above equation for potential difference as below,
\[\Delta V = \dfrac{W}{q}\]
We substitute \[12\,J\] for W and \[{10^{ - 2}}\,C\] for q in the above equation.
\[\Delta V = \dfrac{{12}}{{{{10}^{ - 2}}}}\]
\[ \therefore \Delta V = 1200\,V\]

Therefore, the potential difference between the point a and b is\[\left| {{V_{BA}}} \right| = 1200\,V\].

Additional information:
For the system of two charges, the electric potential energy at a distance r from each other is expressed as,
\[U\left( r \right) = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{{q_1}{q_2}}}{r}\]
The quantity \[\dfrac{q}{{4\pi {\varepsilon _0}r}}\] is known as electric potential of the charge q at a distance r.The relation between electric potential and electric potential energy is,
\[V = \dfrac{U}{q}\]
Here, U is the electric potential energy.
We know that the change in electric potential energy is work done. Therefore, we can write,
\[\Delta V = \dfrac{{\Delta U}}{q} = \dfrac{{dW}}{q}\]
If the unit of charge is coulomb and unit if electric potential energy is joule, then the unit of electric potential difference is volt.

Note:To not get confused between electric potential energy and electric potential. The difference in them is discussed above. If instead of charge, we have given the particle as electron or proton, then the charge of the electron should be taken as \[1.6 \times {10^{ - 19}}\,C\]. We know that electrons and protons have the same magnitude but opposite charges.