Question

Two points charges of $2 \times {10^{ - 7}}C$ and $1.0 \times {10^{ - 7}}$ are 1.0 cm apart. What is the magnitude of the field produced by either charge at the site of the other? Use the standard value of $1/4\pi\in_0$.

Answer 1

Hint:In order to solve this question, we will firstly apply the formula of electric field i.e.$E = \dfrac{{kQ}}{{{r^2}}}$. Then we will substitute two points charges of $2 \times {10^{ - 7}}C$ and $1.0 \times {10^{ - 7}}$to get the required answer.
Formula used-$E = \dfrac{{kQ}}{{{r^2}}}$

Complete step by step answer:

Electric field is the region produced by an electric charge around it whose influence is observed when another charge is brought in that region where the field exists. The force F experienced by electric charge q describes the electric field lines.
The magnitude of the electric field (E) produced by a point charge with a charge of magnitude Q, at a point a distance r away from the point charge, is given by the equation $E = \dfrac{{kQ}}{{{r^2}}}$,
Where k is a constant with a value of $8.99 \times {10^9}N{m^2}/{C^2}$.
The direction of the electric field produced by a point charge is away from the charge if positive, and toward the charge if the charge is negative.
Here we are given that charge 1,${q_1} = 2 \times {10^{ - 7}}C$
Charge 2,${q_2} = 1 \times {10^{ - 7}}C$
Distance between charges, d = 1cm = 0.01m
The magnitude of the field produced by the either charge at the side of the other is given by-
Electric field produced by charge at point A on point B,
$E = \dfrac{1}{{k\pi \in_0}}\dfrac{q}{{{r^2}}}$
$E = 9 \times {10^9} \times \dfrac{{{{10}^{ - 7}}}}{{{{\left( {{{10}^{ - 2}}} \right)}^2}}}$
 $\therefore E = 9 \times {10^6}N{C^{ - 1}}$
Electric field produced by charge at point B on point A,
$E = 9 \times {10^9} \times \dfrac{{2 \times {{10}^{ - 7}}}}{{{{\left( {{{10}^{ - 2}}} \right)}^2}}}$
$\therefore E = 18 \times {10^6}N{C^{ - 1}}$

Note- While solving this question, we should take care that the distance given in centimeters should be converted into meters to get the appropriate answer. Also one can use another formula of electric field i.e.$E = \dfrac{F}{Q}$ to get the required answer.