Electrostatics MCQs for NEET — Physics Questions with Answers

To design a 10 F/700V capacitor using capacitors of rating 10 F/200V, we need to ensure the voltage rating is met by arranging capacitors in series and the capacitance requirement is met by arranging these series combinations in parallel. For the voltage rating: $$ rac{700V}{200V} = 3.5 ightarrow 4 ext{ capacitors in series} $$ Each series combination will have a total capacitance of: $$ rac{10F}{4} = 2.5F $$ To achieve the total capacitance of 10 F, we need: $$ rac{10F}{2.5F} = 4 ext{ such series combinations in parallel} $$ Therefore, the minimum number of capacitors required is: $$ 4 imes 4 = 16 $$ Thus, the answer is 16.

The energy stored in a capacitor is given by the formula $U = rac{1}{2}CV^2$. Initially, the energy stored in the capacitor is $U_1 = rac{1}{2} imes 2 imes 100^2 = 10^4$ J. After changing the capacitance to 10 F, the energy stored is $U_2 = rac{1}{2} imes 10 imes 100^2 = 5 imes 10^4$ J. The change in energy is $ riangle U = U_2 - U_1 = 5 imes 10^4 - 10^4 = 4 imes 10^4$ J.

The maximum capacitance is obtained when all capacitors are connected in parallel. For 10 capacitors each of capacity 5F, the maximum capacitance is $C_{max} = 10 imes 5 = 50$ F. The minimum capacitance is obtained when all capacitors are connected in series. For 10 capacitors each of capacity 5F, the minimum capacitance is $C_{min} = rac{5}{10} = 0.5$ F. The ratio between maximum and minimum capacities is $rac{50}{0.5} = 100:1$.

When 'n' equally spaced plates are connected alternately to form a parallel plate capacitor, the effective number of capacitors in series is (n-1). The total capacitance for capacitors in series is given by the formula: $C_{total} = (n-1) imes x$.

$ V = 4x^2 \Rightarrow \bar E = - { dv \over dr} = -8 x $ Now put the value of x

Knowledge base question

Use equation, $ E = { -dV \over dx } $

When spheres are connected by a conducting wire their potentials become equal $ { C_1 \over C_2 } = { r_1 \over r_2 } = { 1 \over 2 } {q_1 \over q_2 } = { C_1V \over C_2 V } = { C_1 \over C_2 } = {1 \over 2 } $ Now, $ { E_1 \over E_2 } = { { Kq_1 \over r_1^2 } \over {Kq_2 \over r_2^2 } } $ Now find out ratio

$ Q = CV = 5 \mu c $ $ = C^1 = { A\varepsilon_0 \over d-(t -{ t\over k } ) } = { A \varepsilon_0 /d \over 1-({ t -t/k \over d } ) } $ Now put the value