Question

4 charges +q, +2q,+3q and +4q are placed at the corners of the square ABCD of side 0.1 m respectively. The intensity of electric field at the center of the square is found to be $5.1\times {10}^3$ N/C. Find the value of q?

Answer 1

Given,

Side of the square = 0.1 m

The electric field at the center = $5.1\times {10}^3$ N/C

Given situation can be represented as

Distance of the center from each corner will be $\frac{0.1\sqrt{2}}{2}=\frac{0.1}{\sqrt{2}}$

Now we can write

Electric-field due to corner A = $E_A=\frac{kq}{r^2}$

Electric-field due to corner B = $E_B=\frac{k2q}{r^2}$

Electric-field due to corner C = $E_C=\frac{k3q}{r^2}$

Electric-field due to corner D = $E_D=\frac{k4q}{r^2}$

From the given figure we can write the resultant as

Now,

$E_{net}=\sqrt{{\left(\frac{2kq}{r^2}\right)}^2+{\left(\frac{2kq}{r^2}\right)}^2}$

After solving

$E_{net}=2\sqrt{2}\frac{kq}{r^2}=2\sqrt{2}\frac{kq}{\frac{1}{200}}=400\sqrt{2}kq$

According to the question net electric field is $5.1\times {10}^3$ N/C

So, $5.1\times {10}^3=400\times \sqrt{2}\times 9\times {10}^9\times q$

After solving

$q={10}^{-9}C$

Hence the charge is $q={10}^{-9}C$