Question

The electric potential difference between two parallel plates is 2000 volts. If the plates are separated by 2mm, what is the magnitude of electrostatic force (in Newton) on a charge of\[4 \times {10^{ - 6}}C\] located midway between the plates?
A.4N
B.6N
C.8N
D.\[6 \times {10^6}N\]

Answer 1

Hint: The amount of work done in moving a unit positive charge from one point to another is known as electric potential difference. It is denoted by symbol ‘V’ and represented by units ‘Volts’ after the name of Italian Physicist Alessandro Volta.

Formula Used:
Mathematically, the relation between electric field and potential difference is given by:
\[E = \dfrac{{\Delta V}}{d}\]
Where ‘V’ is the electric potential difference, ‘E’ is the electric field and ‘d’ is the distance or separation between the plates.
The electric force is given as,
$F=qE$
Here, $q$ is the charge and $E$ is the electric field.

Complete step by step solution:
Given that the potential difference is \[\Delta V = 2000volts\]
They are separated by a distance, d=2mm
Or it can be written as \[d = 2 \times {10^{ - 3}}m\]
Charge is \[q = 4 \times {10^{ - 6}}C\]
Substituting the values in the above formula and solving, we get
\[E = \dfrac{{2000}}{{2 \times {{10}^{ - 3}}}}\]
On solving the above equation, we get
\[E = {10^6}N/C\]

The electrostatic force that is exerted by the electric field to move the electric charge from one point to another is given by the formula,
$F=qE$
Again substituting all the given values in the above formula and solving for force, we get
$F = 4 \times 10^{ - 6} \times 10^{6}$
$\therefore F=4\,N$
Therefore, the magnitude of the electrostatic force on charge located midway between the plates is 4N.

Hence, Option A is the correct answer.

Note: It is important to note that the direction of the electric field can be known by the direction of movement of charge. If the electric charges are flowing from low potential to high potential, or the field is in this direction, it is taken as positive. On the other hand, if the direction is from high potential to low potential, it is taken as negative.