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Question

# A cavity is made inside a solid conducting sphere and a charge q is placed inside the cavity at the centre. A charge q1 is placed outside the sphere as shown in the figure. Point A is inside the sphere and point B is inside the cavity. Then

A
Electric field at point A is zero.
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B
Electric field at point B is 14πϵ0qy2
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C
Potential at point A is 14πϵ0[q1a+qR]
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D
Potential at point B is 14πϵ0[q1a+qR+qyqr]
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Solution

## The correct options are A Electric field at point A is zero. B Electric field at point B is 14πϵ0qy2 C Potential at point A is 14πϵ0[q1a+qR] D Potential at point B is 14πϵ0[q1a+qR+qy−qr] Charge distribution is as shown. Due to q1, the charge induction is non-uniform but net induced charge due to q1 is zero. At point A field is zero (field inside a conducting sphere) and potential is equal to potential at the centre. VA=VC=14πϵ0qR+14πϵ0q1a At point B, field is due to q only. EB=14πϵ0qy2 The potential at B is due to the charge q at the centre of cavity, −q induced in the surface of cavity and potential at the the centre of sphere. VB=14πϵ0[q1a+qR+qy−qr]

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