V A = V B = V B V A = Q1

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1 lectrostatics hapter : lectrostatics qin. q in Now, q IN for S q q + q q q IN for S + q q q IN for S q + q q q IN for S q. V P V r p.d l r q If is constant, then r p V P V d l r V P V (r p r ) (i ˆ ˆ j). ()i ˆ () ˆ j ( )k ˆ (i ˆ ˆj). i ˆ j ˆ kˆ ( + ) V. The initial potential of the outer shell, K V + K() K() fter connecting the shells, by a wire, the potentials of the shells, V + K(q) and V + K(q) where q is the remnant charge on inner shell. s inner and outer shell are connected, V V q or The later is not possible q K() Thus, V V V So the potential of the outer shell does not change after connecting with wire. () is correct.. ssertion is true, eason is true and eason is a correct explanation for ssertion. V V V V. et us enclose the charge at the mouth of the conical flask with another identical flask. Flux through the closed surface. y symmetry, flux through either flask is. q s both the earthed points are at V, we can redraw the circuit as, and in series q eq V ( V) F ( V) 7. The electric field at the center of the semicircle can be found by calculating the field due to an infinitesimal element and integrating it. Infinitesimal element +q q +q d V Wire with charge density q X

2 Std. XII : Triumph Physics harge on the infinitesimal element (q) dx (d) d.(i) lectric field at O due to this charge (d) k(dq)/.(ii) Where k Substituting (i) in (ii), lectric field d kd/.(iii) X component of electric field (kcos )d d x dcos Y component of the electric field d y dsin (ksin )d (from iii) (from iii) (kcos ) x d Net electric field due to the wire at point O along Y axis (ksin ) y d k () k/ esultant electric field () k/ x y The resultant electric field at the center of the circle 8. onsider the Gaussian surface to be a spherical shell of r negligible thickness at a distance of r from the centre. The net flux through the surface is (r ) The net charge () enclosed dv [ x / 9] x dx v d cos X d sin d where, the limits of x varies from to r. r x net (x )dx 9 r r 9 pplying Gauss s law (r ) net / r r k 9 Hence electric field () at a distance r from r r the centre 9 r r 9 (r r ) 9. The capacitance is to be found between the inner and outer cylinders. k k b a (q) r (q) onsider a Gaussian cylinder of radius r. pplying Gauss s law, (r) q(/) Hence, q/r Potential difference, V dr (q/) ln(b/a) Hence, q [/ln(b/a)]v q net q [/ln(b/a)] V q /V /ln(b/a) Now we can consider the top and bottom parts of the cylinder as two capacitors in parallel. Net capacitance, net + (/) net ( ) l n(b/a) where k and k apacitance of the arrangement (k k ) k ln(b / a) n() l

3 hapter : lectrostatics. onsider a unit charge as shown at P and the co-ordinate frame is chosen as shown in the figure. The force on P acts along the positive y-axis. et us imagine that the P cylinder can be broken into a number of thin disks. Now the y field at P due to one such disk at a distance x from P is x x d x where, x varies from to. Hence, the total field is given by x d d x ( ) x Therefore, electric field at a point at a distance from one end of the cylinder () () 7. The circuit with the switch in ON position is shown in figure (i). We apply the Kirchoff s nd law. onsider the closed loop through the capacitors. Figure (i) Potential drop across the capacitors q q We are traversing the loop from negative to positive. Therefore, potential drop due to battery can be taken as positive. Writing the equation for net potential drop along the loop, q q q q harge flow along the capacitor nergy stored in the capacitor q (/) / In the second case, when the switch is ON, the circuit diagram would be as given in figure (ii) Figure (ii) The circuit is symmetric about. Therefore, we can say that charge entering the capacitor to would be the same as charge leaving through the other capacitor. Therefore, there would be no charge flow along capacitor. Hence, nergy in the capacitor.. The given circuit can be redrawn and reduced to the following: Now, the potential across the two capacitors in parallel in. Hence the charge stored in each is q, q. (i) The other two capacitors are in series. Hence the charge in each of them is q (9/). (ii) (9/) Therefore the potential across the (9/) capacitor is V q /[(9/) ] (/) q

4 Std. XII : Triumph Physics Now working backwards we get the circuit, Hence the circuit gets reduced as shown in the figure below. / / (/) (8/) Since the potential is (/), the charge on the parallel capacitor, q (/).(iii) For the two series capacitors, net (8/) Hence, charge in the capacitors q (8/) The potential across the (8/) capacitor, V (/) We now consider the following circuit: (/) The charge on the X capacitor is (/) (/). et the plates be numbered as shown below. Plates,, and may be treated as a collection of two plates as shown in the diagram. We get five capacitors with top and bottom capacitors having a capacitance / and the rest with capacitance. (/) The equivalent capacitance of the above arrangement ( net ) /7. / /7 V /7 V / /7 V If the potential applied across is V, the charge on the capacitors (q) q V Hence the charges on plate X q (/7) V. The current flow in different segments can be found considering different open loops and applying Kirchoff s junction law. 9V i i V V There are no direct series and parallel connections which can be directly identified. This circuit consists of only resistors. So elements need not be removed from the circuit. et us mark different nodes and loops in the circuit and consider the node. urrent entering the node current leaving the node. urrent enters through, and D. i 7V

5 hapter : lectrostatics et us assume the potential at the nodal point to be V. urrent entering the node i + i + i V 9 V 7 V V 7.(i) urrent leaving the node.(ii) V 7 quating equations (i) and (ii), Voltage at node V 7 urrent flowing through wire (i ) V 9.7. Net charge inside the sphere dv Due to spherical symmetry, we get, r rdr r( r)dr. lectric flux through: i. X-Y plane ii. Y-Z plane 8 8 iii. X-Z plane Hence, the required ratio : 8 : 7. Potential of the bigger drop / n potential on each droplet / V 8. edistribution of charges takes place and during flow of charges some energy is lost as heat. 9. V (V )dv V [ + 8] [ + 8] 8 J. V, k d V d s d increases, decreases Hence V increases.

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