Question

Charge 2Q and -Q are placed as shown in figure.
The point at which electric field intensity is zero will be somewhere:

• A. Between -Q and 2Q
• B. On the left of -Q
• C. On the right of 2Q
• D. On the perpendicular bisector of line joining the charges

Answer 1

The correct option is B On the left of -Q

R.E.F image

Solution It will be left to -Q as there

the direction of electric failed due

to -Q and 2Q are opposite and

the separation due Q is letter then

that of 2Q as to equalize the

magnitude of fields.

$F_1=\frac{kd}{d^2},f_2=\frac{2KQ}{(d+l{)}^2}$

option - B is correct