Q)A hollow charged metal sphere has radius r. If the potential differe...
Q)A hollow charged metal sphere has radius r. If the potential differe...
Solution:
Given,
Potential difference (V) between the surface of the hollow charged metal sphere and a point at a distance 3r from the centre.
Radius of the metal sphere (r).
To find:
Electric field intensity at a distance 3r from the centre.
As we know that,
Electric field intensity is the negative gradient of the potential difference.
So,
Electric field intensity (E) = - dV/dr
Where,
V = potential difference
r = distance from the centre of sphere.
Now,
Let's consider a Gaussian surface of radius 3r, enclosing the hollow metal sphere.
Using Gauss's law,
Electric flux through the Gaussian surface is given by,
Electric flux (Φ) = (Qenclosed) / ε0
Where,
Qenclosed = charge enclosed by the Gaussian surface
ε0 = permittivity of free space.
As the metal sphere is hollow, the charge enclosed by the Gaussian surface is zero.
Therefore,
Electric flux (Φ) = 0.
Now,
Let's find the electric field intensity at a distance 3r from the centre using the potential difference.
We know that,
Potential difference (V) between the surface of the hollow charged metal sphere and a point at a distance 3r from the centre is given by,
V = (Q / 4πε0r) - (Q / 4πε0(3r))
Where,
Q = charge on the hollow metal sphere.
Simplifying the above equation, we get
V = Q / 4πε0r (1/3 - 1)
V = - Q / 4πε0r (2/3)
V = - (2/3) (Q / 4πε0r)
Now,
Electric field intensity (E) at a distance 3r from the centre is given by,
E = - dV/dr
Differentiating V with respect to r, we get
dV/dr = (2/3) (Q / 4πε0r^2)
Putting the value of Q from the above equation, we get
dV/dr = (2/3) (V / r)
Therefore,
Electric field intensity at a distance 3r from the centre is given by,
E = - dV/dr = - (2/3) (V / r) = - (2/3) (V / 3r) = - V / (3r)
Hence,
Electric field intensity at a distance 3r from the centre is V/6r.
Therefore, the correct option is (2) V/6r.
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