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Question: Two charges of \(+10\mu C\) and \(+20\mu C\) are separated by a distance of \(2\text{ cm}\). The net...

Two charges of +10μC+10\mu C and +20μC+20\mu C are separated by a distance of 2 cm2\text{ cm}. The net potential (electric) due to the pair at the middle point of the line joining the two charges is:

Explanation

Solution

Two solve these types of questions, we need to know how to find the electric potential due a single charge at a particular point. Then we will find the effect of electric potential at the middle point of the line joining the two charges due to the two positive charges of the given magnitude in the question.

Complete step-by-step solution:
It is given that two charges of +10μC+10\mu C and +20μC+20\mu C are separated by a distance of 2 cm2\text{ cm} and we need to find the net potential (electric) due to the pair at the middle point of the line joining the two charges. Since the distance between the two charges is 2 cm2\text{ cm}, hence the middle point of the line joining the two charges will be at a distance of 1 cm1\text{ cm} from both the charges. We know that the electric potential due to a positive charge at a particular distance is given by the following formula:
V=kqrV=\dfrac{kq}{r}
Hence the electric potential at a distance of 1 cm1\text{ cm} due to the +10μC+10\mu C charge will be:
V1=(9×109)×(10×10−6)0.01{{V}_{1}}=\dfrac{\left( 9\times {{10}^{9}} \right)\times \left( 10\times {{10}^{-6}} \right)}{0.01}
And the electric potential at a distance of 1 cm1\text{ cm} due to the +20μC+20\mu C charge will be:
V2=(9×109)×(20×10−6)0.01{{V}_{2}}=\dfrac{\left( 9\times {{10}^{9}} \right)\times \left( 20\times {{10}^{-6}} \right)}{0.01}
Thus, the electric potential at the middle point of the line joining the two charges i.e., at a distance of 1 cm1\text{ cm} due to both the charges will be:
V=(9×109)×(10×10−6)0.01+(9×109)×(20×10−6)0.01 ⇒V=(9×109)×(10×10−6+20×10−6)0.01 ⇒V=(9×109)×(30×10−6)0.01 ⇒V=27×106 V ∴V=27 MV \begin{aligned} & V=\dfrac{\left( 9\times {{10}^{9}} \right)\times \left( 10\times {{10}^{-6}} \right)}{0.01}+\dfrac{\left( 9\times {{10}^{9}} \right)\times \left( 20\times {{10}^{-6}} \right)}{0.01} \\\ & \Rightarrow V=\dfrac{\left( 9\times {{10}^{9}} \right)\times \left( 10\times {{10}^{-6}}+20\times {{10}^{-6}} \right)}{0.01} \\\ & \Rightarrow V=\dfrac{\left( 9\times {{10}^{9}} \right)\times \left( 30\times {{10}^{-6}} \right)}{0.01} \\\ & \Rightarrow V=27\times {{10}^{6}}\text{ V} \\\ & \therefore V=27\text{ MV} \\\ \end{aligned}
Thus, the electric potential at the middle point of the line joining the two charges will be 27 MV27\text{ MV}.

Note: The electric potential can be defined as the work done in moving a unit charge from one point to another. For a positive charge the value of electric potential is positive while for a negative charge the value of electric potential is negative.