# Solved Problem-7

Problem-7

A pair of conducting planes meets at an angle of 60. A point charge +

*Q*is located at a distance â€˜*a*â€™ from both the planes. Find the electric field intensity induced at the foot of the perpendicular.Solution

For

*Ï†**=*60, number of charges is,The image configuration is shown in the Figure.
We have to find the intensity at the point
Now,
Also,

Also,
field intensity at

(a) Due to charges at
[Since charges at

(b) Due to charges at
[Since charges at

(c) Due to charges at
Hence, the total field intensity is given as,

*P*.**Image configuration**

*P*:(a) Due to charges at

*B*and*E*is,*B*and*C*are equal and opposite, their horizontal components cancel each other and the resultant field intensity is in the vertical direction.](b) Due to charges at

*C*and*D*is,*C*and*D*are equal and opposite, their horizontal components of field intensity cancel each other, but vertical components give the resultant field intensity.](c) Due to charges at

*A*and*F*is,