Question

Two point charges of \[ + 5 \times {10^{ - 19}}\;{\rm{C}}\] and \[ + 20 \times {10^{ - 19}}\;{\rm{C}}\] are separated by a distance of \[2\;{\rm{m}}\]. Find the point on the line joining them at which electric field intensity is zero.

Answer 1

Hint: The above problem can be resolved using the concepts and the mathematical formula for the electric field at any point. The electric field in any region is that particular space at which the electrostatic force is experienced on either charge. If the net electric field at any point is zero, then we are supposed to find the distance from that particular point. To calculate this, we must know that the net electric field can be zero only when the electric field, due to two charges, has its sum of zero.

Complete step by step answer:
Given:
The magnitude of first charge is \[{q_1} = + 5 \times {10^{ - 19}}\;{\rm{C}}\].
The magnitude of the second charge is \[{q_2} = + 20 \times {10^{ - 19}}\;{\rm{C}}\].
The separation of charges is \[r = 2\;{\rm{m}}\].
We know the expression for the electric field is given by,
\[E = \dfrac{{kq}}{{{r^2}}}\]
Here, k is the coulomb’s constant.
Let us assume that P is any point such that the charge \[{q_1}\] is x distance from p and the charge \[{x_2}\] be \[\left( {2 - x} \right)\] distance from \[{q_2}\]. And at this point the net electric field is zero.
As, the net intensity of the electric field is zero, then the electric fields of both the charges are equal.
\[\begin{array}{l}
{E_1} = {E_2}\\
\dfrac{{k{q_1}}}{{{x^2}}} = \dfrac{{k{q_2}}}{{{{\left( {2 - x} \right)}^2}}}
\end{array}\]
On solving the above equation and by substituting the values, we get,
\[\begin{array}{l}
\dfrac{{k{q_1}}}{{{x^2}}} = \dfrac{{k{q_2}}}{{{{\left( {2 - x} \right)}^2}}}\\
\dfrac{{5 \times {{10}^{ - 19}}\;{\rm{C}}}}{{{x^2}}} = \dfrac{{20 \times {{10}^{ - 19}}\;{\rm{C}}}}{{{{\left( {2 - x} \right)}^2}}}\\
15{x^2} + 20x - 20 = 0
\end{array}\]
On further solving the above quadratic equation as,
\[\begin{array}{l}
15{x^2} + 20x - 20 = 0\\
\Rightarrow 3{x^2} + 4x - 4 = 0\\
\Rightarrow x = 0.67\;{\rm{m}}
\end{array}\]
Therefore, the point on the line joining them at which electric field intensity is zero is of \[0.67\;{\rm{m}}\] from the charge \[{q_1}\].

Note:To resolve the given problem, the concept and application of the electric field should be cleared. One must know that the net electric field is zero only when the field due to the two entities has its sum equivalent to zero.