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A thin metallic spherical shell of radius R carries a charge Q on its surface. A point charge Q/2 is placed at the centre c and another charge +2Q is placed outside the shell at a distance x from the centre as shown,find 1)the force on the charge at the centre of shell and at the point A, 2)the electric flux through the cell

A thin metallic spherical shell of radius R carries a charge Q on its surface. A point charge Q/2 is placed at the centre c and another charge +2Q is placed outside the shell at a distance x from the centre as shown,find 
1)the force on the charge at the centre of shell and at the point A,
2)the electric flux through the cell

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Grade:12

2 Answers

Arun
25757 Points
4 years ago
Dear Akshat
Since, the electric field inside a spherical shell is zero, the force on the charge placedat the centre of the shell is zero. For the charge at A, the shell will behave as if theentire charge ‘Q’ is placed at the centre of the shell. Therefore, the total charge is
Q + Q/2 = 3Q/2
Since its distance from 2Q is x
then electric field at A is – 
E = k (3Q/2) / x^2
so electric force F = 2Q * E = k * 3 Q^2 / x^2
Now
Since, the total charge enclosed by the shell is qen = Q/2
,the total flux according to Gauss’s law is
phi = Q/2 / epsilon = Q / 2 * epsilon
Kushagra Madhukar
askIITians Faculty 628 Points
2 years ago
Dear student,
Please find the answer to your question below.
 
Since, the electric field inside a spherical shell is zero, the force on the charge placedat the centre of the shell is zero. For the charge at A, the shell will behave as if theentire charge ‘Q’ is placed at the centre of the shell. Therefore, the total charge is
Q + Q/2 = 3Q/2
Since its distance from 2Q is x
then electric field at A is – 
E = k (3Q/2) / x^2
so electric force F = 2Q * E = k * 3 Q^2 / x^2
Now
Since, the total charge enclosed by the shell is qen = Q/2
,the total flux according to Gauss’s law is
phi = Q/2 / epsilon = Q / 2 * epsilon
 
Thanks and regards,
Kushagra

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