electro stat test sol'ns

8/2/2019 Electro Stat Test Sol'Ns

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Ans.1 Clearly, the capacitors are in parallel. Therefore,

Ctot = C1 + C2 = 4µF + 6µF = 10µF

Charge on C1 = 4 x 10-6 F x 20 V = 8.0 x 10-5 C = 80 µC.

Charge on C2 = 6 x 10-6 F x 20 V = 12.0 x 10-5 C = 120 µC.

Total charge = 48 µC + 72 µC = 200 µC.

Ans.2 The capacitors in series are connected one after the other in a row, back to back. Incase of series capacitors, positive plate of one capacitor is attached to the negative plate of the successive capacitor. The distribution of charges from the battery results in equal chargesappearing on each one of the capacitors connected in series. Also note that the sum of thepotential difference across each capacitor is equal to the total potential difference. Therefore,V1

+V2 = V,which can be written in terms of charge as

A singleequivalent capacitance in series would imply

This further gives,

Ans.3 Here capacitors of 20 μf and 20 μf are joined in parallel, combined capacitanceisC’=20+20 =40 μf. As C and C’ are in series, their resultant capacitance is 30 μf, we have



Ans.4 The given circuit can be redrawn as

It has been clear that capacitors form wheatstone bridge arrangement.

As

, hence bridge is balanced. (The point R and S are at the same potential)

As no charge can accumulate on , which thus become ineffective.

Between P and Q two series combinations ( ) and ( ) are connected in parallel. If C’and C’’ are the equivalent capacitances of these combinations then,



Ans.5 In case of parallel arrangement, one end of all capacitors is attached to the same

junction and the other end of all capacitors is connected to each other at another junction. The

potential difference is the same across each of them.

The charge on each capacitor will not necessarily be the same if they have differentcapacitances because while the charging of the capacitors, the charges developed across the

capacitor plates will be proportional to the capacitance of each capacitor since q=C V and V isthe same for capacitors in parallel.

Now, for two capacitors in parallel, the total charge Q stored across an equivalent capacitor isdivided between the two capacitors. Let the charges be Q1 and Q2.Thus

Q = Q1 + Q2

Then we can write

But and since the potential difference or voltage V is the same across eachcapacitor.

Hence, the equivalent capacitance of a set of capacitors in parallel is the sum of the individualcapacitances.

Ans. 6

let’s identify which capacitors are in series and which are in parallel.C2 and C3 are in series.

Their effective capacitance is

Or So the circuit is rearranged and the equivalent network looks like,



Then, C1 and C23 are in parallel.

, net capacitance = C123 =C1 + C23 = 200 pF

C4 is now in series with C123.So the net equivalent capacitance will be,

Ans 7. Here Q=20 μC=

a. Electric Field at the centre of the sphere is zero, because intensity at any point inside a hollow

sphere is always zero.

b. Electric Field on the outer surface of the sphere is-

Ans. 8 Let us suppose that q is an isolated point charge situated at centre of sphere of radius

According to Coulomb’s law electric field intensity at every point on the surface of the sphere is

where is a unit vector directed from +q to the surface element.



Consider a small area element dS represented by vector

Where is a unit vector along outdrawn normal to the area element.

Integrating over closed surface area of the sphere, we get

Hence

If there are point charges, , each will contribute to the electric flux.

Therefore =

=

=

where Q=



Hence total electric flux over the closed surface S in vacuum is 1/ times the total charge (Q)contained inside S.

electro stat test sol'ns

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