A 400 pF capacitor is charged with a 100 V battery. After disconnecting the battery this capacitor is connected with another 400 pF capacitor. Then the energy loss is:
1. \( 1 \times 10^{-6}\text{ J} \)
2. \( 2 \times 10^{-6}\text{ J} \)
3. \( 3 \times 10^{-6}\text{ J} \)
4. \( 4 \times 10^{-6}\text{ J} \)
View Answer
Energy loss when connecting two capacitors is \( \Delta U = \frac{1}{2} \frac{C_1 C_2}{C_1 + C_2} (V_1 - V_2)^2 \). Here, \( C_1 = C_2 = 400\text{ pF} \), \( V_1 = 100\text{ V} \), and \( V_2 = 0 \). This gives \( \Delta U = \frac{1}{4} C V^2 = 1 \times 10^{-6}\text{ J} \).
Energy per unit volume for a capacitor having area A and separation d kept at potential difference V is given by:
1. \( \frac{1}{2} \varepsilon_0 \frac{V^2}{d^2} \)
2. \( \frac{1}{2 \varepsilon_0} \frac{V^2}{d^2} \)
3. \( \frac{\varepsilon_0 V^2 A^2}{2d^2} \)
4. \( \frac{1}{2} \frac{V^2 A^2}{\varepsilon_0 d^2} \)
View Answer
Energy density (energy per unit volume) of a capacitor is given by the formula \( u = \frac{1}{2} \varepsilon_0 E^2 \). Substituting electric field \( E = \frac{V}{d} \) gives \( u = \frac{1}{2} \varepsilon_0 \frac{V^2}{d^2} \).
The distance between the plates of an isolated charged parallel plate condenser is \(4 \text{ mm}\) and potential difference is \(60 \text{ volts}\). If the distance between the plates is increased to \(12 \text{ mm}\), then
1. The potential difference of the condenser will become \(180 \text{ volts}\)
2. The P.D. will become \(20 \text{ volts}\)
3. The P.D. will remain unchanged
4. The charge on condenser will reduce to one third
View Answer
For an isolated capacitor, the charge \(Q\) remains constant. \(Q = C_1 V_1 = C_2 V_2\). Since \(C = \frac{\epsilon_0 A}{d}\), we have \( \frac{\epsilon_0 A}{d_1} V_1 = \frac{\epsilon_0 A}{d_2} V_2 \implies \frac{V_1}{d_1} = \frac{V_2}{d_2} \). Thus, \(V_2 = V_1 \frac{d_2}{d_1} = 60 \text{ V} \times \frac{12 \text{ mm}}{4 \text{ mm}} = 180 \text{ V}\).
A capacitor of capacitance \(C\) is charged to a potential \(V\). The flux of the electric field through a closed surface enclosing the capacitor is
1. \(\frac{CV}{2\epsilon_0}\)\( \)
2. \(\frac{2CV}{\epsilon_0}\)\( \)
3. \(\frac{CV}{\epsilon_0}\)\( \)
4. zero
View Answer
A capacitor consists of two plates with equal and opposite charges, \(+Q\) and \(-Q\), where \(Q = CV\). If a closed surface encloses the entire capacitor, the net charge enclosed within the surface is \(Q_{\text{enclosed}} = (+Q) + (-Q) = 0\). By Gauss's Law, the total electric flux through this closed surface is \(Phi_E = \frac{Q_{\text{enclosed}}}{\epsilon_0}\). Therefore, \(Phi_E = 0\).
Two vertical metallic plates carrying equal and opposite charges are kept parallel to each other like a parallel plate capacitor. A small spherical metallic ball is suspended by a long insulated thread such that it hangs freely in the centre of the two metallic plates. The ball, which is uncharged, is taken slowly towards the positively charged plate and is made to touch that plate. Then the ball will
1. stick to the positively charged plate
2. come back to its original position and will remain there
3. oscillate between the two plates touching each plate in turn
4. oscillate between the two plates without touching them
View Answer
Initially, the uncharged ball is attracted to the positive plate by induction. Upon touching, it acquires positive charge and is repelled by the positive plate, moving towards the negative plate. Upon touching the negative plate, it acquires negative charge and is repelled by the negative plate, moving back to the positive plate. This continuous charge transfer and repulsion causes the ball to oscillate between the plates, touching each in turn.
Choose the CORRECT statement :
1. C will increase on increasing Q
2. C will increase on decreasing Q
3. C will increase on decreasing V
4. C doesnot depend on Q & V
View Answer
The capacitance of a capacitor is a constant determined by its geometric shape, size, and the medium between the plates. It is independent of the charge \( Q \) and potential difference \( V \) applied to it.
A charge of \( + 2.0 \times 10^{-8}\text{ C} \) is placed on the positive plate and a charge of \( -1.0 \times 10^{-8}\text{ C} \) on the negative plate of a parallel-plate capacitor of capacitance \( 1.2 \times 10^{-3}\ \mu\text{F} \). Calculate the potential difference developed between the plates.
1. \( 25\text{ V} \)
2. \( 7.5\text{ V} \)
3. \( 12.5\text{ V} \)
4. \( 50\text{ V} \)
View Answer
The potential difference depends on the charge on the inner facing surfaces, which is given by \( q = \frac{q_1 - q_2}{2} = \frac{2.0 \times 10^{-8} - (-1.0 \times 10^{-8})}{2} = 1.5 \times 10^{-8}\text{ C} \). Using \( V = \frac{q}{C} \), we get \( V = \frac{1.5 \times 10^{-8}}{1.2 \times 10^{-9}} = 12.5\text{ V} \).
From a supply of identical capacitors rated \( 8\ \mu\text{F}\ \), \( 250\text{V} \), the minimum number required to form a composite \( 16\ \mu\text{F}\ \), \( 1000\text{V} \) capacitor is
1. \( 2 \)
2. \( 4 \)
3. \( 8 \)
4. \( 32 \)
View Answer
To handle \( 1000\text{V} \), \( N_s = \frac{1000}{250} = 4 \) capacitors must be connected in series. The capacitance of one row is \( C_s = \frac{8}{4} = 2\ \mu\text{F} \). To get a total capacitance of \( 16\ \mu\text{F} \), we need \( N_p = \frac{16}{2} = 8 \) parallel rows. Total number \( = N_s \times N_p = 4 \times 8 = 32 \).