Electrostatic Potential and Capacitance Test

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Electrostatic Potential and Capacitance Test - 72

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Question 1
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Two parallel plate capacitors of capacitances $$C$$ and $$2C$$ are connected in parallel and charged to a potential difference $$V$$. The battery is then disconnected and the region between the plates of the capacitor $$C$$ is completely filled with a material of dielectric constant $$K$$. The potential difference across the capacitors now becomes :

A
$$\dfrac {3V}{K+2}$$

B
$$KV$$

C
$$\dfrac {V}{K}$$

D
$$\dfrac {3}{KV}$$

Solution

Initial total charge of the system is: $$Q_i=Q_1+Q_2=CV+2CV=3CV$$
When dielectric is inserted in $$C$$ so the capacitance becomes $$KC$$
Final charge, $$Q_f=Q'_1+Q'_2=KC V'+2CV'=(K+2)CV'$$ where $$V'=$$ common potential after disconnected the battery.
As battery is disconnected so total charge will remain unchanged.
Thus, $$Q_i=Q_f \Rightarrow 3CV=(K+2)CV'$$
$$\therefore V'=\dfrac{3V}{K+2}$$
Question 2
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A series combination of $$n_1$$ capacitors, each of value $$C_1$$, is charged by a source of potential difference 4 V. When another parallel combination of $$n_2$$ capacitors, each of value $$C_2$$, is charged by a source of potential difference V, it has the same (total) energy stored in it, as the first combination has. The value of $$C_2$$, in terms of $$C_1$$, is then

A
$$\dfrac {2C_1}{n_1n_2}$$

B
$$16\dfrac {n_2}{n_1}C_1$$

C
$$2\dfrac {n_2}{n_1}C_1$$

D
$$\dfrac {16C_1}{n_1n_2}$$

Solution

In series, the equivalent capacitance is: $$\displaystyle C_S=\frac{C_1}{n_1}$$ and
Energy stored, $$\displaystyle E_S=\frac{1}{2}C_SV_S^2=\frac{1}{2}(C_1/n_1)(4V)^2=\frac{8C_1V^2}{n_1}$$

In parallel, the equivalent capacitance is $$\displaystyle C_P={n_2C_2}$$ and
Energy stored , $$\displaystyle E_P=\frac{1}{2}C_PV_P^2=\frac{1}{2}(n_2C_2)(V)^2=\frac{n_2C_2V^2}{2}$$

Here, $$E_S=E_P$$

$$\displaystyle \frac{8C_1V^2}{n_1}=\frac{n_2C_2V^2}{2}$$

$$\displaystyle \therefore C_2=\frac{16C_1}{n_1n_2}$$
Question 3
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A parallel plate capacitor is connected to a battery. The quantities charge, voltage, electric field and energy associated with this capacitor are given by $$Q_0, V_0, E_0$$, and $$U_0$$ respectively. A dielectric slab is now introduced to fill the space between the plates with the battery still in connection. The corresponding quantities now given by $$Q, V, E$$ and $$U$$ are related to the previous ones are :

A
$$Q > Q_0$$

B
$$V > V_0$$

C
$$E > E_0$$

D
$$U < U_0$$

Solution

As the battery till in connection so potential will remained unchanged. i,e $$V=V_0$$
$$E_0=\dfrac{V_0}{d} $$ and $$E=\dfrac{V}{d}=\dfrac{V_0}{d} \Rightarrow E=E_0$$ where $$d=$$ separation of plates.
Let initial capacitance $$=C_0$$
When dielectric slab with dielectric constant K is inserted, the capacitance becomes $$C=KC_0$$.
Thus, $$Q_0=C_0V_0$$ and $$Q=CV=KC_0V_0=KQ_0 \Rightarrow Q>Q_0$$ as $$(K>1)$$
Also, $$U_0=(1/2)C_0V_0^2$$ and $$U=(1/2)CV^2=(1/2)KC_0.V_0=KU_0$$
Hence, $$U>U_0$$
Question 4
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The equivalent capacitance between x and y is :

A
$$5/6 \mu F$$

B
$$7/6 \mu F$$

C
$$8/3 \mu F$$

D
$$1\mu F$$

Solution

The equivalent is shown in the figure below.
Here, $$C_{eq1}=\dfrac{(1+1)\times 1}{(1+1)+1}=2/3 \mu F$$
and $$C_{XY}=C_{eq1}+2=(2/3)+2=8/3 \mu F$$
Question 5
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Find equivalent capacitance between points A and B :
[Assume each conducting plate is having same dimensions and neglect the thickness of the plate, $$\dfrac {\varepsilon_0A}{d}=7\mu F$$, where A is area of plates]

A
$$7\mu F$$

B
$$11\mu F$$

C
$$12\mu F$$

D
$$15\mu F$$

Solution

Here, $$\displaystyle C=\frac{A\epsilon_0}{d}=7\mu F$$. The equivalent circuits are shown in the figure.
Now, $$\displaystyle C_{eq}=C+\frac{C(C/2)}{C+(C/2)}=C+(1/3)C=(4/3)C$$
and $$\displaystyle C_{AB}=C+\frac{C_{eq}C}{C_{eq}+C}=C+\frac{(4/3)C C}{(4/3)C+C}=C+(4/7)C=(11/7)C=(11/7)(7)=11 \mu F$$
Question 6
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Figure shows a parallel plate capacitor with plate area $$A$$ and plate separation $$d$$. A potential difference is being applied between the plates. The battery is then disconnected and a dielectric slab of dielectric constant $$K$$ is placed in between the plates of the capacitor as shown.
Now, answer the following questions based on above information. The electric field in the gaps between the plates and the dielectric slab will is :

A
$$\dfrac {\varepsilon AV}{d}$$

B
$$\dfrac {V}{d}$$

C
$$\dfrac {KV}{d}$$

D
$$\dfrac {V}{d-t}$$

Solution

The gap between the plates and the dielectric field is filled with air/vacuum. So electric field here is same as electric field with vacuum between the parallel plate capacitor, i,e, $$E_o=V/d$$
Question 7
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A parallel plate capacitor of plate area $$A$$ and plate separation $$d$$ is charged to potential difference $$V$$ and then the battery is disconnected. A slab of dielectric constant $$K$$ is then inserted between the plates of capacitor so as to fill the space between the plates. If $$Q, E$$ and $$W$$ denote respectively, the magnitude of charge on each plate electric field between the plates (after the slab is inserted), and work done on the system, in question, in the process of inserting the slab, then which is wrong?

A
$$Q=\dfrac {\epsilon_0AV}{d}$$

B
$$Q=\dfrac {\epsilon_0KAV}{d}$$

C
$$E=\dfrac {V}{Kd}$$

D
$$W=\dfrac {\epsilon_0AV^2}{2d}\left (1-\dfrac {1}{K}\right )$$

Solution

Initial capacitance, $$\displaystyle C_0=\dfrac{A\epsilon_0}{d}$$
So initial charge on the capacitor is: $$Q_0=C_0V=\dfrac{AV\epsilon_0}{d}$$
As the battery is disconnected so charge remains unchanged.
So, $$\displaystyle Q=Q_0=\dfrac{AV\epsilon_0}{d}$$
After insert the dielectric the capacitance becomes, $$C'=KC_0$$ where K dielectric constant.
Now potential becomes, $$V'=Q/C'=Q_0/KC_0=V/K$$
Field, $$E=V'/d=\dfrac{V}{Kd}$$
Initial energy, $$\displaystyle U_0=\frac{1}{2}C_0V^2=\frac{AV^2\epsilon_0}{2d}$$ and
final energy $$\displaystyle U_f=\frac{1}{2}C'V'^2=(1/2)(KC_0)(V/K)^2=\frac{AV^2\epsilon_0}{2Kd}$$

Since, work done $$=$$Decrease in Potential Energy
Work done, $$\displaystyle W=-\Delta U=-(U_f-U_0)=U_0-U_f=\frac{1}{2} \frac{AV^2\epsilon_0}{d}(1-1/K)$$
Question 8
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Positive charge Q is uniformly distributed throughout the volume of a dielectric sphere of radius R. A point mass having charge +q and mass m is fired towards the centre of the sphere with velocity v from a point A at distance r(r> R) from the centre of the sphere. Find the minimum velocity v so that it can penetrate R/2 distance of the sphere. Neglect any resistance other than electric interaction. Charge on the small mass remains constant throughout the motion.

A
$$\displaystyle \left[\frac{1}{2 \pi \varepsilon_0} \frac{Qq}{Rm} \left(\frac{r-R}{r} + \frac{3}{4}\right) \right]^{1/2}$$

B
$$\displaystyle \left[\frac{1}{2 \pi \varepsilon_0} \frac{Qq}{Rm} \left(\frac{r-R}{r} + \frac{3}{8}\right) \right]^{1/2}$$

C
$$\displaystyle \left[\frac{1}{2 \pi \varepsilon_0} \frac{Qq}{Rm} \left(\frac{r-R}{r} - \frac{3}{8}\right) \right]^{1/2}$$

D
$$\displaystyle \left[\frac{1}{4 \pi \varepsilon_0} \frac{Qq}{Rm} \left(\frac{r-R}{r} + \frac{3}{4}\right) \right]^{1/2}$$

Solution

Initial energy are kinetic and potential energy given by-

$$K_i=\dfrac{1}{2}mv^2$$ and $$U_i=\dfrac{Qq}{4\pi\epsilon_o r}$$

Finally at last point , its velocity get reduced to $$0$$, and potential at a point inside sphere is given by-

$$V=\dfrac{Q}{4\pi\epsilon_o}\dfrac{3R^2-r^2}{2R^3}$$

At $$r=\dfrac{R}{2}$$, $$U_f=qV$$

$$\implies U_f=\dfrac{Qq}{4\pi\epsilon_o}\dfrac{3R^2-\dfrac{R^2}{4}}{2R^3}$$

$$\implies U_f=\dfrac{11}{8}\dfrac{Qq}{4\pi\epsilon_oR}$$

Now applying conservation of mechanical energy-

$$K_i+U_i=K_f+U_f$$

$$\implies \dfrac{1}{2}mv^2+\dfrac{Qq}{4\pi\epsilon_o r}=0+\dfrac{11}{8}\dfrac{Qq}{4\pi\epsilon_o R}$$

$$\implies mv^2=\dfrac{Qq}{2\pi\epsilon_o}\left(\dfrac{11}{8R}-\dfrac{1}{r}\right)$$

$$\implies v^2=\dfrac{Qq}{2\pi\epsilon_oRm}\left(\dfrac{11}{8}-\dfrac{R}{r}\right)$$

$$\implies v^2=\dfrac{Qq}{2\pi\epsilon_oRm}\left(1+\dfrac{3}{8}-\dfrac{R}{r}\right)$$

$$\implies v^2=\dfrac{1}{2\pi\epsilon_o}\dfrac{Qq}{Rm}\left(\dfrac{r-R}{r}+\dfrac{3}{8}\right)$$

$$\implies v=\sqrt{\dfrac{1}{2\pi\epsilon_o}\dfrac{Qq}{Rm}\left(\dfrac{r-R}{r}+\dfrac{3}{8}\right)}$$

Answer-(B)
Question 9
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A parallel plate capacitor of area A and separation $$d$$ is charged to potential difference $$V$$ and removed from the charging source A dielectric slab of constant $$K = 5$$, thickness $$d$$ and area $$\displaystyle \frac{A}{3}$$ is inserted as shown in the figure Let $$\displaystyle \sigma _{1}$$ be free charge density at the conductor-dielectric surface and $$\displaystyle \sigma _{2}$$ be the charge density at the conductor-vacuum surface :

A
the electric field have the same value inside the dielectric as in the free space between the plates

B
the ratio $$\displaystyle \frac{\sigma _{1}}{\sigma _{2}}$$ is equal to $$\displaystyle \frac{1}{5}$$

C
the new capacitance is $$\displaystyle \frac{7\varepsilon _{0}A}{3d}$$

D
the new potential difference is $$\displaystyle \frac{3}{7}V $$

Solution

Potential for each plate remain same over whole are if potential difference thems is way V then V = Ed i.e. E is also same inside the plates
To Keep E same free change dansity is changed i.e. charge redistributes itself
To find new capacitance, two capacitors can be taken as connected in parallel Then
$$\displaystyle C_{eq}=\frac{5.\epsilon _{0}A/3}{d}+\frac{\epsilon _{0}.2A/3}{d}=\frac{7\epsilon _{0}A}{3d}$$
Q = CV, as Q remains unchanged V is changed to $$\displaystyle \frac{3}{7}V$$
Question 10
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Directions For Questions

The figure shows four parallel plate capacitors : A, B, C and D. Each capacitor carries the same charge $$q$$ and has the same plate area $$A$$. As suggested by the figure, the plates of capacitors A and C are separated by a distance $$d$$ while those of B and D are separated by a distance $$2d$$. Capacitors A and B are maintained in vacuum while capacitors C and D contain dielectrics with constant $$K=5$$.

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Which list below places the capacitors in order of increasing capacitance ?

A
A, B, C, D

B
B, A, C, D

C
B, A, D, C

D
A, B, D, C

Solution

For A: the capacitance $$C_A=\dfrac{A\epsilon_0}{d}$$
For B: the capacitance $$C_B=\dfrac{A\epsilon_0}{2d}$$
For C: the capacitance $$C_C=\dfrac{Ak\epsilon_0}{d}=\dfrac{5A\epsilon_0}{d}$$
For D: the capacitance $$C_D=\dfrac{Ak\epsilon_0}{2d}=\dfrac{5A\epsilon_0}{2d}$$
so, $$C_B<C_A<C_D<C_C$$