Question

Charges $20,30,-40$ and $50\mu C$ are at the corners of a square of $10cm$. The field at the point of intersection of the diagonals is :

• A. $360\sqrt{10}\times {10}^5\frac{N}{C}$
• B. $360\times {10}^5\frac{N}{C}$
• C. $360\times {10}^6\frac{N}{C}$
• D. $36\sqrt{10}\times {10}^5\frac{N}{C}$

Answer 1

The correct option is A $360\sqrt{10}\times {10}^5\frac{N}{C}$

The side length is 10cm so diagonal length will be $10\sqrt{2}.$
The field at point of intersection of diagonals due to charge $20\mu C$ is

$=\frac{9\times {10}^9\times (20\times {10}^{-6})}{(0.1/\sqrt{2}{)}^2}$ away from it.

The field at point of intersection of diagonals due to charge $-40\mu C$ is

$=\frac{9\times {10}^9\times (40\times {10}^{-6})}{(0.1/\sqrt{2}{)}^2}$ towards it.

Hence field due to these towards the $-40\mu C$ charge is $1080\times {10}^{5}N/C$

The field at point of intersection of diagonals due to charge $30\mu C$ is

$=\frac{9\times {10}^9\times (30\times {10}^{-6})}{(0.1/\sqrt{2}{)}^2}$ away from it.

The field at point of intersection of diagonals due to charge $50\mu C$ is

$=\frac{9\times {10}^9\times (50\times {10}^{-6})}{(0.1/\sqrt{2}{)}^2}$ away from it.

Hence, net field towards $50\mu C$ charge is $360\times {10}^{5}N/C$

Hence, the net field at the point $=360\sqrt{10}\times {10}^{5}N/C$