**Three concentric spherical shells have radii**

(1)

(1)

*a*,

*b*and

*c*(

*a*<

*b*<

*c*) and have surface charge densities σ, − σ and σ respectively. If

*V*

_{A},

*V*

_{B }and

*V*

_{C}denote the potentials of the three shells, then for

*c*=

*a + b*, we have

(1) *V*_{C} = *V*_{B} ≠ *V*_{A}

(2) *V*_{C} ≠ *V*_{B} ≠ *V*_{A}

(3) *V*_{C }= *V*_{B} = *V*_{A}

(4) *V*_{C} = *V*_{A} ≠ *V*_{B}

The potential *V *of a spherical shell of radius *r* having surface charge density σ is given by

*V* = (1/4πε_{0})(4π*r*^{2}σ/*r*) = σ*r/*ε_{0} where ε_{0} is the permittivity of free space.

Potential *V*_{A} of the shell A is given by

*V*_{A} = σ*a/*ε_{0} – σ*b/*ε_{0} + σ*c/*ε_{0} = **(σ /ε_{0})[c – (b – a)]**

Potential *V*_{B} of the shell B is given by

*V*_{B} = – σ*b/*ε_{0} + (1/4πε_{0})(4π*a*^{2}σ/*b*) + σ*c/*ε_{0}

Or, *V*_{B} = **σ /ε_{0} [c – (b^{2 }– a^{2})/b]**

Potential *V*_{C }of the shell C is given by

*V*_{C} = σ*c/*ε_{0 }– (1/4πε_{0})(4π*b*^{2}σ/*c*) + (1/4πε_{0})(4π*a*^{2}σ/*c*)

Or, *V*_{C} = σ*/*ε_{0} [*c *– (*b*^{2 }–* a*^{2})/*c*] = σ*/*ε_{0} [*c *– (*b* + *a*)(*b*–* a*)/*c*]

Since *c* = *a + b* we obtain

*V*_{C} = **σ /ε_{0} [c – (b– a)]**

The correct option therefore is* V*

**.**

_{C}=*V*_{A}≠*V*_{B}**(2)** Three capacitors each of capacitance *C* and of breakdown voltage *V* are joined in series. The capacitance and breakdown voltage of the combination will be

(1) 3 *C*, *V*/3

(2) *C*/3, 3 *V*

(3) 3 *C*, 3*V*

(4) *C/*3, *V/*3

This is a very simple question. The effective calacitance *C*_{eff}* *of the combination of three capacitors in *series *is given by the reciprocal relation,

1/*C*_{eff}* = *1*/C*_{1} +1*/C*_{2 }+1*/C*_{3}

Here *C*_{1} = *C*_{2} = *C*_{3} = *C* so that *C*_{eff} = *C/*3

The breakdown voltage* V*_{eff} for the series combination is the sum of the individual breakdown voltages:

*V*_{eff} = **3 V**

**(3)** The electric potential at a point (x, y, z) is given by *V* = − x^{2}y − xz^{3} + 4. The electric field E at that point is:

(1)* ***E** = **i** 2xy + **j** (x^{2} + y^{2}) + **k** (3xz − y^{2})

(2)** ****E** = **i** z ^{3 }+ **j** xyz + **k** z^{2}

(3)** ****E** = **i** (2xy − z^{3}) + **j** xy^{2} + **k** 3z^{2}x

(4)** ****E** = **i** (2xy + z^{3}) + **j** x^{2} + **k** 3xz^{2}

The electric field **E** is the negative gradient of potential:

**E** = − ∂*V/*∂*r* = − (**i** ∂*V/*∂x + **j** ∂*V/*∂y + **k**** **∂*V/*∂z)

This gives **E** = **i **(2xy + z^{3}) + **j** x^{2} + **k** 3xz^{2} as given in option (4).