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Referring to the spherical equipotential...

Referring to the spherical equipotential lines in (Fig. 3.36)A, find
(i) `vec E = f ( r)`
(ii) `vec E` - pattern.
a. .
b. .
c. .

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(a) For the first equipotential
`60 V xx (10)/(100) (m) = 6 Vm`
For the second equipotential
`30 V xx (20)/(100) (m) = 6 Vm`
For the third equipotential
`10 V xx (60)/(100) (m) = 6 Vm`
We can understand that are product of potential (V) and radial distance (r) of equipotential lines [ as in Fig 3.36 B] is equal to `6 Vm^-1`. Hence, general relation of potential (V) and radial distance (r) can be written as
`V = (6)/ ( r) Vm^-1`
Since `vec E - (del V)//(del r) hat r`, substituting `V = 6//r`, we have
`vec E = - (del)/(del r) (- (6)/( r)) hat r Vm^-1`
or `vec E = (6)/(r^2) hat r Vm^-1`
(b) Since `vec E` is directed in `hat r` - direction and obeys inverse square law, E must be outward [as shown in Fig. 3.36 C].
The above E - field must be caused by a positive point charge.
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