**Charges +q, +q, –q and –q coulomb are arranged at the corners of an imaginary rectangle of sides 6m and 4m, as shown in the adjoining figure. A and B are the mid points of the longer sides. The potential difference between A and B is (in volts)**

(a) 2.4×10

(a) 2.4×10

^{6}q (b) 4.8×10^{6}q (c) 2.4×10^{9}q (d) 4.8×10^{9}q (e) 9.6×10^{9}qThe potential at A due to the charges +q and +q is 2×(1/4πε

Similarly, the resultant potential at B is 2×(1/4πε

_{0})(q/3) and the potential at the same point due to the charges –q and –q is 2×(1/4πε_{0}))(–q/5). The resultant potential at A is 2×(1/4πε_{0})[(q/3)–(q/5)] = 2×(1/4πε_{0}) (2q/15)= (1/4πε_{0}) (4q/15).Similarly, the resultant potential at B is 2×(1/4πε

_{0})[(q/5) – (q/3)] =2×(1/4πε

The potential difference between points A and B is (1/4πε

The following question appeared in Punjab CEET 2003 question paper:

_{0}) (–2q/15) = (1/4πε_{0})(–4q/15).The potential difference between points A and B is (1/4πε

_{0}))[(4q/15) – (–4q/15)] = (1/4πε_{0})×(8q/15). You should remember that 1/4πε_{0}= 9×10^{9}. Therefore, the P.D. is 9×10^{9}×(8/15)q = 4.8×10^{9}q volts.The following question appeared in Punjab CEET 2003 question paper:

**As shown in figure, if the point C is earthed and the point A is given a potential of 2000 volt, then the potential at B will be****(a) 400 V (b) 500 V (c) 1000 V (d) 1300 V**

The effective capacitance between the points B and C is 15 μF because 10μF and 10μF in series in the upper branch make 5μF and this is in parallel with 10μF in the lower branch. So, in effect 2000 V is applied across the series combination of 5μF and 15μF connected between A and C. Since the potential difference across capacitors connected in series is inversely proportional to the capacitance, the applied P.D. of 2000 volts get divided in the ratio 3: 1 across 5μF and 15μF [1500 V : 500V]. The potential at B with respect to the grounded point C is therefore 500V.

Now consider the following question which appeared in IIT 2002 question paper:

On connecting the capacitors (in parallel), the net capacitance of the system is 2C and the net voltage across the combination is ½ (V

Therefore, final energy of the system is(½) × 2C[(V

On subtracting the final energy from the initial energy, we obtain the decrease in energy as ¼ C(V

**Two identical capacitors have the same capacitance C. One of them is charged to potential V**

(a) ¼ C(V

The initial energy of the system is ½ CV_{1}and the other to potential V_{2}. The negative ends of the capacitors are connected together. When the positive ends are also connected, the decrease in energy of the combined system is(a) ¼ C(V

_{1}^{2}– V_{2}^{2}) (b) ¼ C(V_{1}^{2}+ V_{2}^{2}) (c) ¼ C(V_{1}– V_{2})^{2}(d) ¼ C(V_{1}+ V_{2})^{2}_{1}^{2}+ ½ CV_{2}^{2}= ½ C(V_{1}^{2}+V_{2}^{2}).On connecting the capacitors (in parallel), the net capacitance of the system is 2C and the net voltage across the combination is ½ (V

_{1}+V_{2}). [The total charge is C(V_{1}+V_{2}) so that the net voltage is C(V1+V2)/ 2C = (V1+V2)/2].Therefore, final energy of the system is(½) × 2C[(V

_{1}+V_{2})/2]^{2}= ¼ C(V_{1}^{2}+V_{2}^{2}+2V_{1}V_{2}).On subtracting the final energy from the initial energy, we obtain the decrease in energy as ¼ C(V

_{1}–V_{2})^{2}, given by option (c).You will find

**with solution***more multiple choice questions on Electrostatics***.***in the post dated 9th October 2006*
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