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Solved Problem-7

A pair of conducting planes meets at an angle of 60. A point charge +Q is located at a distance ‘a’ from both the planes. Find the electric field intensity induced at the foot of the perpendicular.
For φ = 60, number of charges is,
Description: 167836.png
The image configuration is shown in the Figure.
We have to find the intensity at the point P.
Description: 167825.png
Also, Description: 167818.png
Description: 167809.png
Image configuration
Description: 167802.png
Description: 167791.png
Also, Description: 167782.png
Description: 167773.png
field intensity at P:

(a) Due to charges at B and E is,
Description: 219593.png
[Since charges at 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,
Description: 167759.png
[Since charges at 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,
Description: 167750.png
Hence, the total field intensity is given as,
Description: 167742.png

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