The following two questions from Electrostatics were included under Straight Objective Type (Multiple choice, single answer type) in the IIT-JEE 2008 question paper:

*q*/3,

*q*/3 and –2

*q*/3 placed at points A, B and C respectively as shown in the figure. Take O to be the centre of the circle of radius

*R*and angle CAB = 60º

(a) The electric field at point O is* q*/8πε_{0}*R*^{2}

(b) The potential energy of the system is zero

(c) The magnitude of the force between the charges at C and B is *q*^{2}/54πε_{0}*R*^{2}

(d) The potential at point O is *q*/12πε_{0}*R*

The electric field at O is due to the negative charge at C only since the equal positive charges situated at A and B will produce equal and opposite fields at O (and they will mutually cancel). The field at O is therefore negative and the option (a) is obviously wrong. Option (b) also is obviously wrong.

The magnitude of the force (*F*) between the charges at C and B is given by

*F =* (1/4πε_{0}) [(2*q*/3)(*q*/3)/(2*R*sin60º)^{2}].

Thus *F* = *q*^{2}/54πε_{0}*R*^{2} as given in option (c).

(2) A parallel plate capacitor with plates of unit area and separation *d* is filled with a liquid of dielectric constant *K* = 2. The level of liquid is *d*/3 initially. Suppose the liquid level decreases at a constant speed *V*. The time constant as a function of time *t* is

(a) 6ε_{0}*R/*(5*d *+ 3*Vt*)

(b) (15d + 9Vt) ε_{0}*R**/*(2*d*^{2} – 3*dVt *– 9*V*^{2}*t*^{2})

(c) 6ε_{0}*R/*(5*d * – 3*Vt*)

(d) (15d – 9Vt) ε_{0}*R**/*(2*d*^{2} + 3*dVt *– 9*V*^{2}*t*^{2})

When a dielectric slab of thickness *t*_{1} and dielectric constant *K *is introduced in between the plates of a parallel plate air capacitor of plate area *A* and plate separation *d*, the effective capacitance changes from ε_{0}*A/d* to ε_{0}*A/*[*d*–*t*_{1} + (*t*_{1}*/K*)]

The effective capacitance *C* of a parallel plate capacitor with parallel dielectric slabs of thickness *t*_{1}, *t*_{2},* t*_{3} etc. of dielectric constants *K*_{1}, *K*_{2},* K*_{3} etc. respectively is given by the series combined value of capacitors with these dielectrics. Therefore,

1/*C* = 1/(*K*_{1}ε_{0}*A/t*_{1}) + 1/(*K*_{2}ε_{0}*A/t*_{2}) + 1/(*K*_{3}ε_{0}*A/t*_{3}) + etc.

With a single slab of thickness *t*_{1}* *and dielectric constant* K* introduced between the plates, we have

1/*C* = 1/(*K*ε_{0}*A/t*_{1}) + 1/[ε_{0}*A/*(*d* – *t*_{1})] from which *C *= ε_{0}*A/*[*d*–*t*_{1} + (*t*_{1}*/K*)]

In the present problem, the thickness of the layer of liquid (which serves as the dielectric slab) at the instant *t* is (*d*/3) – *V t* and the thickness of the air space is (2*d/*3) +* V t.*

The effective capacitance is therefore given by

*C *= ε_{0}*A/*[{(2*d/*3) +* Vt*} + {(*d*/3) – *Vt*}*/k*)]

Substituting for *A *= 1 and *K* = 2, we obtain *C* = 6ε_{0}*/*(5*d *+ 3*Vt*).

The ** time constant** of the circuit =

**=**

*RC***6**

**ε**.

_{0}*R/*(5*d*+ 3*Vt*)`Reality is merely an illusion, albeit a very persistent one. `

–Albert Einstein

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