Question

A positive charge +Q is located at 5cm on the positive y-axis. A negative charge -Q is located at 5cm on the positive x-axis. A positive test charge is at the origin where it feels a force "F" from the positive charge above it. What is the net force from both charges?

• A. $2F$
• B. $F^2$
• C. $\frac{F}{2}$
• D. $\sqrt{2}F$

Answer 1

The correct option is D $\sqrt{2}F$

The force on test positive charge at the origin due to positive charge on the y-axis is acting downwards and due to the negative charge on the x-axis is acting towards the right as shown in the figure. Here the magnitude of both forces are equal that is equal to $F$because $F=\frac{kQq}{r}$ and both case $r$ and $Qq$ are common.

Thus, net force, $R=\sqrt{F^2+F^2}=\sqrt{2}F$