(i) Explain the following statements giving reason:
(I) An equipotential surface through a point is normal to the electric field at that point.
(II) When a dielectric is placed in an external electric field, the electric field inside the dielectric is less than that outside it.
(III) The potential difference between the plates of a charged parallel plate capacitor decreases when its plates are brought closer.
(ii) Obtain an expression for the work done to dissociate the system of three charges \(q\), \(-4q\) and \(2q\) placed at the vertices A, B and C respectively of an equilateral triangle of side 'a'.
(i) Explain the following statements giving reason:
(I) An equipotential surface through a point is normal to the electric field at that point.
(II) When a dielectric is placed in an external electric field, the electric field inside the dielectric is less than that outside it.
(III) The potential difference between the plates of a charged parallel plate capacitor decreases when its plates are brought closer.
(ii) Obtain an expression for the work done to dissociate the system of three charges \(q\), \(-4q\) and \(2q\) placed at the vertices A, B and C respectively of an equilateral triangle of side 'a'.
Show Hint
Solution and Explanation
On an equipotential surface the potential is the same at every point. So the work done in moving a test charge \(q_0\) between two points of the surface is zero.
Take a small displacement \(d\vec{l}\) along the surface. Then \(W = q_0 \vec{E}\cdot d\vec{l} = q_0 E\, dl \cos\theta = 0\).
Since \(q_0\), \(E\) and \(dl\) are not zero, \(\cos\theta = 0\), so \(\theta = 90^{\circ}\). The field is normal to the surface.
If \(E\) had a component along the surface, work would be needed to move a charge on it, and the surface would not be equipotential.
Step 2: Statement (II): field inside a dielectric is smaller:
A dielectric has no free charges, but its molecules get polarised in an external field \(\vec{E}_0\). The positive and negative charges shift slightly in opposite directions.
This leaves a bound charge on the two faces of the slab. The bound charges make an induced field \(\vec{E}_p\) inside the slab, opposite to \(\vec{E}_0\).
\[ E = E_0 - E_p < E_0 \]
So the net field inside is less than the field outside. In fact \(E = E_0/K\), where \(K\) is the dielectric constant.
Step 3: Statement (III): potential difference falls when plates come closer:
A charged capacitor that is not connected to a battery keeps its charge \(Q\) constant.
Its capacitance is \(C = \dfrac{\varepsilon_0 A}{d}\). When the separation \(d\) decreases, \(C\) increases.
\[ V = \frac{Q}{C} = \frac{Qd}{\varepsilon_0 A} \]
With \(Q\) fixed, \(V\) is proportional to \(d\). So the potential difference decreases when the plates are brought closer.
Step 4: Part (ii): set up the three-charge system:
Place \(q\) at A, \(-4q\) at B and \(2q\) at C of an equilateral triangle of side \(a\). Every pair of charges is a distance \(a\) apart.
The work needed to dissociate the system (take all charges to infinity) equals the negative of the potential energy stored in the system.
Step 5: Potential energy of the system:
\[ U = \frac{1}{4\pi\varepsilon_0}\left[\frac{q(-4q)}{a} + \frac{(-4q)(2q)}{a} + \frac{(2q)(q)}{a}\right] \]
\[ U = \frac{1}{4\pi\varepsilon_0 a}\left[-4q^2 - 8q^2 + 2q^2\right] = -\frac{10q^2}{4\pi\varepsilon_0 a} \]
Step 6: Work done to dissociate:
\[ W = -U = \frac{10q^2}{4\pi\varepsilon_0 a} = \frac{5q^2}{2\pi\varepsilon_0 a} \]
The work is positive, because the system is bound (U is negative) and energy must be supplied to pull the charges apart.
Final Answer:
(I) \(E\) is normal to an equipotential surface because no work is done along it. (II) Bound charges reduce the field inside a dielectric. (III) \(V = Q/C\) falls as \(d\) falls at constant \(Q\). The work needed to dissociate the three charges is given below. \[ \boxed{W = \frac{10q^2}{4\pi\varepsilon_0 a}} \]
Top CBSE CLASS XII Physics Questions
- A boy of mass 50 kg is standing at one end of a, boat of length 9 m and mass 400 kg. He runs to the other, end. The distance through which the centre of mass of the boat boy system moves is
- A capillary tube of radius r is dipped inside a large vessel of water. The mass of water raised above water level is M. If the radius of capillary is doubled, the mass of water inside capillary will be
- A circular disc is rotating about its own axis. An external opposing torque 0.02 Nm is applied on the disc by which it comes rest in 5 seconds. The initial angular momentum of disc is
- A circular disc is rotating about its own axis at uniform angular velocity \(\omega.\) The disc is subjected to uniform angular retardation by which its angular velocity is decreased to \(\frac {\omega}{2}\) during 120 rotations. The number of rotations further made by it before coming to rest is
- A constant power is supplied to a rotating disc. The relationship between the angular velocity $\omega$ of the disc and number of rotations (n) made by the disc is governed by
Top CBSE CLASS XII capacitor with a dielectric Questions
- A parallel plate capacitor with plate area \( A \) and plate separation \( d \) has a capacitance \( C_0 \). A slab of dielectric constant \( K \) having area \( A \) and thickness \( \left(\frac{d}{4}\right) \) is inserted in the capacitor, parallel to the plates. Find the new value of its capacitance.
Two air-filled capacitors of capacitances $C_1$ and $C_2$ are connected in parallel with a dc battery. After the capacitors are fully charged, a slab of dielectric constant $K$ is inserted between the plates of each capacitor. How will the (i) charge on each capacitor and (ii) energy stored in the capacitors be affected after the slab is introduced?
Two air-filled capacitors of capacitances $C_1$ and $C_2$ are connected in parallel with a dc battery. After the capacitors are fully charged, a slab of dielectric constant $K$ is inserted between the plates of each capacitor. How will the (i) charge on each capacitor and (ii) energy stored in the capacitors be affected after the slab is introduced?
- Two air-filled capacitors of capacitances $C_1$ and $C_2$ are connected in parallel with a dc battery. After the capacitors are fully charged, a slab of dielectric constant K is inserted between the plates of each capacitor. How will the (i) charge on each capacitor and (ii) energy stored in the capacitor affected after the slab is introduced.
A parallel plate capacitor of capacitance $C$ is charged to $V$ volt by a battery. After sometime the battery is disconnected and the distance between the plates is doubled. A slab of dielectric constant $k = 1.8$ is then introduced to completely fill the space between the plates. How will the following be affected?
(a) The capacitance of the capacitor.
(b) The electric field between the plates of the capacitor.
(c) The energy stored in the capacitor.Justify your answer in each case.
Top CBSE CLASS XII Questions
- 2, b, c are in A.P. and the range of determinant $\begin{vmatrix}1&1&1\\ 2&b&c\\ 4&b^{2}&c^{2}\end{vmatrix}$ is [2, 16]. Then find range of c is
- A boy of mass 50 kg is standing at one end of a, boat of length 9 m and mass 400 kg. He runs to the other, end. The distance through which the centre of mass of the boat boy system moves is
- A capillary tube of radius r is dipped inside a large vessel of water. The mass of water raised above water level is M. If the radius of capillary is doubled, the mass of water inside capillary will be
- A circular disc is rotating about its own axis. An external opposing torque 0.02 Nm is applied on the disc by which it comes rest in 5 seconds. The initial angular momentum of disc is
- A circular disc is rotating about its own axis at uniform angular velocity \(\omega.\) The disc is subjected to uniform angular retardation by which its angular velocity is decreased to \(\frac {\omega}{2}\) during 120 rotations. The number of rotations further made by it before coming to rest is