Energy Stored & Dielectrics — MH-CET Physics MCQs with Solutions
Free MH-CET Physics Energy Stored & Dielectrics MCQs with step-by-step solutions (20 questions). Part of Electrostatics. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Energy Stored & Dielectrics · easy · theory
The energy stored in a charged capacitor is given by:
A. $QV$
B. $CV^2$
C. $\dfrac{1}{2}CV^2 = \dfrac{Q^2}{2C} = \dfrac{1}{2}QV$ ✓ Correct
D. $\dfrac{Q^2}{C}$
Solution: The factor of one-half arises because the potential difference builds up from zero to $V$ as the capacitor charges.
Q2 — Energy Stored & Dielectrics · medium · theory
The energy density of an electrostatic field $E$ in vacuum is:
A. $\dfrac{1}{2}\varepsilon_0E^2$ ✓ Correct
B. $\dfrac{E^2}{2\varepsilon_0}$
C. $\varepsilon_0E^2$
D. $\dfrac{1}{2}\varepsilon_0E$
Solution: The energy of a charged capacitor can be regarded as residing in the field between the plates, at this energy per unit volume.
Q3 — Energy Stored & Dielectrics · easy · theory
The dielectric constant of a material is defined as the ratio of:
A. The capacitance with the dielectric to that with vacuum ✓ Correct
B. The energy stored to the volume
C. The field with the dielectric to that without
D. The charge to the potential difference
Solution: $K = \dfrac{C}{C_0}$, a pure number always greater than one for any real dielectric.
Q4 — Energy Stored & Dielectrics · medium · theory
When a dielectric is placed in an external electric field, the field inside the dielectric is:
A. Reduced, because polarisation charges oppose the applied field ✓ Correct
B. Unchanged
C. Reduced to exactly zero
D. Increased by the polarisation
Solution: Induced surface charges create an opposing field, so the net internal field becomes $\dfrac{E_0}{K}$.
Q5 — Energy Stored & Dielectrics · hard · theory
A charged parallel plate capacitor is disconnected from the battery and the plate separation is then doubled. The stored energy will:
A. Double ✓ Correct
B. Halve
C. Quadruple
D. Remain unchanged
Solution: The charge is fixed, and $U = \dfrac{Q^2}{2C}$ with $C$ halved gives twice the energy — supplied by the work done in pulling the plates apart.
Q6 — Energy Stored & Dielectrics · medium · theory
A dielectric is inserted into a capacitor while the battery remains connected. The charge on the plates:
A. Decreases
B. Remains unchanged
C. Becomes zero
D. Increases, since the capacitance rises at constant voltage ✓ Correct
Solution: With $V$ fixed by the battery and $C$ increased $K$ times, $Q = CV$ increases in the same proportion.
Q7 — Energy Stored & Dielectrics · hard · theory
When a charged capacitor is connected to an uncharged one, the total stored energy:
A. Remains exactly the same
B. Decreases, the loss appearing as heat in the connecting wires ✓ Correct
C. Becomes zero
D. Increases
Solution: Charge is conserved but energy is not: the redistribution current dissipates energy resistively.
Q8 — Energy Stored & Dielectrics · medium · theory
The dielectric strength of an insulating material is the:
A. Ratio of its capacitance to that of vacuum
B. Energy stored per unit area
C. Charge it can store per unit volume
D. Maximum field it can withstand without breaking down ✓ Correct
Solution: Beyond this field the material ionises and begins to conduct, which sets the voltage rating of a capacitor.
Q9 — Energy Stored & Dielectrics · medium · numerical
A capacitor of $20\,\mu\text{F}$ is charged to $500\text{ V}$. The energy stored is:
A. $5.0\text{ J}$
B. $10.0\text{ J}$
C. $2.5\text{ J}$ ✓ Correct
D. $0.25\text{ J}$
Solution: $U = \dfrac{1}{2}CV^2 = \dfrac{1}{2} \times 20 \times 10^{-6} \times (500)^2 = 2.5\text{ J}$.
Q10 — Energy Stored & Dielectrics · medium · numerical
A capacitor of $10\,\mu\text{F}$ is charged to $200\text{ V}$. The energy stored is:
A. $2\text{ J}$
B. $0.1\text{ J}$
C. $0.4\text{ J}$
D. $0.2\text{ J}$ ✓ Correct
Solution: $U = \dfrac{1}{2}(10 \times 10^{-6})(4 \times 10^4) = 0.2\text{ J}$.
Q11 — Energy Stored & Dielectrics · hard · numerical
A capacitor carries a charge of $10^{-3}\text{ C}$ and has a capacitance of $2\,\mu\text{F}$. The energy stored is:
A. $0.5\text{ J}$
B. $0.25\text{ J}$ ✓ Correct
C. $2.5\text{ J}$
D. $0.025\text{ J}$
Solution: $U = \dfrac{Q^2}{2C} = \dfrac{(10^{-3})^2}{2 \times 2 \times 10^{-6}} = \dfrac{10^{-6}}{4 \times 10^{-6}} = 0.25\text{ J}$.
Q12 — Energy Stored & Dielectrics · hard · numerical
The energy density in a region where the electric field is $1000\text{ V/m}$ is:
A. $8.85 \times 10^{-6}\text{ J/m}^3$
B. $1.77 \times 10^{-5}\text{ J/m}^3$
C. $4.43 \times 10^{-6}\text{ J/m}^3$ ✓ Correct
D. $4.43 \times 10^{-9}\text{ J/m}^3$
Solution: $u = \dfrac{1}{2}\varepsilon_0E^2 = \dfrac{1}{2} \times 8.85 \times 10^{-12} \times 10^6 \approx 4.43 \times 10^{-6}\text{ J/m}^3$.
Q13 — Energy Stored & Dielectrics · easy · numerical
If the potential difference across a capacitor is doubled, the energy stored becomes:
A. Half
B. Unchanged
C. Twice
D. Four times ✓ Correct
Solution: $U = \dfrac{1}{2}CV^2 \propto V^2$, so doubling $V$ quadruples the energy.
Q14 — Energy Stored & Dielectrics · medium · numerical
A capacitor of $4\,\mu\text{F}$ is charged to $50\text{ V}$. The energy stored is:
A. $10^{-2}\text{ J}$
B. $10^{-4}\text{ J}$
C. $5 \times 10^{-3}\text{ J}$ ✓ Correct
D. $2.5 \times 10^{-3}\text{ J}$
Solution: $U = \dfrac{1}{2}(4 \times 10^{-6})(2500) = 5 \times 10^{-3}\text{ J}$.
Q15 — Energy Stored & Dielectrics · hard · numerical
A $2\,\mu\text{F}$ capacitor charged to $100\text{ V}$ is connected across an uncharged $2\,\mu\text{F}$ capacitor. The energy lost is:
A. $10^{-2}\text{ J}$
B. $5 \times 10^{-3}\text{ J}$ ✓ Correct
C. Zero
D. $2.5 \times 10^{-3}\text{ J}$
Solution: Initial energy $= \dfrac{1}{2}(2 \times 10^{-6})(10^4) = 10^{-2}\text{ J}$. The common potential is $50\text{ V}$, so the final energy is $\dfrac{1}{2}(4 \times 10^{-6})(2500) = 5 \times 10^{-3}\text{ J}$. The loss is $5 \times 10^{-3}\text{ J}$.
Q16 — Energy Stored & Dielectrics · hard · numerical
A dielectric of constant $5$ is inserted into a capacitor with the battery still connected. The stored energy becomes:
A. Unchanged
B. Five times as large ✓ Correct
C. Twenty-five times as large
D. One-fifth as large
Solution: At constant $V$, $U = \dfrac{1}{2}CV^2 \propto C$, and the capacitance rises five-fold.
Q17 — Energy Stored & Dielectrics · medium · numerical
A capacitor holds a charge of $2 \times 10^{-4}\text{ C}$ at a potential difference of $100\text{ V}$. The energy stored is:
A. $10^{-2}\text{ J}$ ✓ Correct
B. $5 \times 10^{-3}\text{ J}$
C. $2 \times 10^{-2}\text{ J}$
D. $2 \times 10^{-6}\text{ J}$
Solution: $U = \dfrac{1}{2}QV = \dfrac{1}{2} \times 2 \times 10^{-4} \times 100 = 10^{-2}\text{ J}$.
Q18 — Energy Stored & Dielectrics · hard · numerical
The energy density in a region where the electric field is $2000\text{ V/m}$ is:
A. $3.54 \times 10^{-5}\text{ J/m}^3$
B. $4.43 \times 10^{-6}\text{ J/m}^3$
C. $1.77 \times 10^{-5}\text{ J/m}^3$ ✓ Correct
D. $8.85 \times 10^{-6}\text{ J/m}^3$
Solution: $u = \dfrac{1}{2} \times 8.85 \times 10^{-12} \times 4 \times 10^6 = 1.77 \times 10^{-5}\text{ J/m}^3$.
Q19 — Energy Stored & Dielectrics · medium · numerical
A capacitor of $1\,\mu\text{F}$ is charged to $1000\text{ V}$. The energy stored is:
A. $0.5\text{ J}$ ✓ Correct
B. $1\text{ J}$
C. $0.05\text{ J}$
D. $5\text{ J}$
Solution: $U = \dfrac{1}{2}(10^{-6})(10^6) = 0.5\text{ J}$.
Q20 — Energy Stored & Dielectrics · medium · numerical
If the charge on an isolated capacitor is doubled, the energy stored becomes:
A. Unchanged
B. Four times ✓ Correct
C. Half
D. Twice
Solution: With $C$ fixed, $U = \dfrac{Q^2}{2C} \propto Q^2$, so doubling the charge quadruples the energy.