The following four questions which appeared in the All India Engineering/Architecture Entrance Examination 2009 (AIEEE 2009) will be beneficial to most of the entrance test takers. Those who prepare for AP Physics C exam and physics GRE may take special note of question no.2. Here are the questions with solution:

**(1)** A charge *Q* is placed at each of the opposite corners of a square. A charge *q* is placed at each of the other two corners. If the net electric force on Q is zero, then *Q/q* equals

(1) – 1/√2

(2) – 2√2

(4) 1

The electric force F_{QQ} (Fig.) on *Q* because of the charge Q at the opposite corner is *repulsive* and is given by

F_{1} = (1/4πε_{0})(*Q*^{2}_{ }/2*a*^{2}) since the distance between the charges *Q *and *Q* is 2×*a/*√2* = a*√2

The electric forces *F*_{Qq} and *F*_{Qq} (Fig.) on *Q* because of the charges *q* and *q *at the adjacent corners are perpendicular to each other. Each force has magnitude given by

*F*_{Qq} = (1/4πε_{0})(*Qq/a*^{2}).

Their resultant F_{2} has magnitude √2 times the above value:

*F*_{2} = (1/4πε_{0})(√2 *Qq/a*^{2}).

Since the net force on *Q* is zero, we have

F_{1} + F_{2} = 0

Therefore, (1/4πε_{0})(*Q*^{2}_{ }/2*a*^{2}) + (1/4πε_{0})(√2 *Qq/a*^{2}) = 0

This gives *Q/q* = – 2**√2**

[The charges *Q *and *q* must be of *opposite* sign so that the force between *Q *and *q* is attractive to ensure that F_{1} and F_{2} are opposite in directions and the net force on *Q* is zero as given in the question].

**(2) **Let *P*(*r*) = *Qr*/π*R*^{4} be the charge density distribution for a solid sphere of radius *R* and total charge *Q*. For a point ‘p’ inside the sphere at distance *r*_{1} from the centre of the sphere, the magnitude of electric field is

(1) *Qr*_{1}^{2}* /*3πε_{0 }*R*^{4}

(2) 0

(3) *Q/*4πε_{0 }*r*_{1}^{2}

(4) *Qr*_{1}^{2}* /*4πε_{0 }*R*^{4}

Charge *dQ *contained in the spherical shell of radius *r* and thickness *dr* is (*Qr*/π*R*^{4})(4π*r*^{2}*dr*) = (4*Q/**R*^{4})* r*^{3}*dr*

Charge *Q*_{1}* *contained in the spherical volume of radius *r*_{1} is therefore given by

*Q*_{1} = _{0} ∫^{r}^{1} (4*Q/**R*^{4})* r*^{3}*dr* = (4*Q/**R*^{4})(*r*_{1}^{4}/4) = *Qr*_{1}^{4}* /**R*^{4}

If E is the electric field at distance *r*_{1} from the centre of the sphere, we have from Gauss theorem,

4π*r*_{1}^{2}E = *Qr*_{1}^{4}* /**R*^{4}ε_{0} from which

**E =** *Qr*_{1}^{2}* /***4πε _{0 }R^{4}**

[Remember that the charges distributed with spherical symmetry outside the point p will produce zero field at p and the field at p is indeed given correctly by the above expression].

**(3)** Two points P and Q are maintained at the potentials of 10 V and – 4 V, respectively. The work done in moving 100 electrons from P to Q is

(1) 2.24×10^{–16} J

(2) – 9.60×10^{–17} J

(3) 9.60×10^{–17} J

(4) – 2.24×10^{–16} J

This is a simple question. The potential difference ∆V between P and Q is 10 V – (– 4 V) = 14 V. Since the electrons are negatively charged, external work (positive work) has to be done to move them from the higher potential point P to the lower potential point Q.

The work done = q∆V = (100×1.6×10^{–19})×14 J = **2.24****×10 ^{–16} J**

**(4)** *This question contains Statement-1 and statement-2. Of the four choices given after the statements, choose the one that best describes the two statements.*

Statement 1 :** **For a charged particle moving from point *P *to point *Q*, the net work done by an electrostatic field on the particle is independent of the path connecting point *P *to point *Q*.

Statement 2 :** **The net work done by a conservative force on an object moving along a closed loop is zero.

(1) Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation of Statment-1.

(2) Statment-1 is true, Statement-2 is true; Statement-2 is not the correct explanation of Statement-1.

(3) Statement-1 is false, Statement-2 is true.

(4) Statement-1 is true, Statement-2 is false

Electrostatic field is conservative and the net work done by an electrostatic field on a charged particle is dependent only on the initial and final positions of the charged particle. (It is independent of the path connecting the initial and final positions).

The correct option is (1).

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