Electrostatic Potential and Capacitance Test

Result

Electrostatic Potential and Capacitance Test - 67

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Weekly Quiz Competition

Question 1
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The equivalent capacitance between points A and B is?

A
$$C/4$$

B
$$C/2$$

C
$$C$$

D
$$2C$$

Solution
Question 2
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A dielectric slab is inserted between the plates of a capacitor. The charge on the capacitor is $$Q$$ and the magnitude of the induced charge on each surface of the dielectric is $$Q'$$.

A
$$Q'$$ must be smaller than $$Q$$

B
$$Q'$$ must be larger than $$Q$$

C
$$Q'$$ must be equal to $$Q$$.

D
None

Solution

The relation between the induced charge and the charge of the capacitor is given as:
$$Q'=Q\left(1-\dfrac1K\right)$$

The dielectric constant is always greater than one. So, the value of the induced cvharge will always be less than Q.

Option $$D$$ is correct.
Question 3
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In the given figure, the equivalent capacity between A and B is?

A
$$3C$$

B
$$C/3$$

C
$$3C/2$$

D
Infinity

Solution
Question 4
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In the given arrangement of capacitors, equivalent capacitance between points M and N is?

A
$$\dfrac{5C_0}{4}$$

B
$$\dfrac{3C_0}{4}$$

C
$$\dfrac{4C_0}{5}$$

D
None of these

Solution
Question 5
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In a region electric field is parallel to x-axis. The equation of equipotential surface is

A
$$y = C$$

B
$$x = C$$

C
$$z = C$$

D
None of these

Solution
Question 6
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Directions For Questions

Consider Fig.

...view full instructions

The charge appearing on $$C_2$$ is?

A
$$E\left(\dfrac{C_3C_4}{C_1+C_2}\right)$$

B
$$E\left(\dfrac{C_1C_2}{C_1+C_2}\right)$$

C
$$E\left(\dfrac{C_1C_2}{C_3+C_4}\right)$$

D
$$E\left(\dfrac{C_3C_4}{C_3+C_4}\right)$$

Solution
Question 7
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Directions For Questions

Each capacitor has capacitance C in the figure shown below.

...view full instructions

The equivalent capacitance between $$4$$ and $$5$$ is?

A
$$\left(\dfrac{3C}{4}\right)$$

B
$$\left(\dfrac{3C}{2}\right)$$

C
$$\left(\dfrac{3C}{5}\right)$$

D
$$\left(\dfrac{5C}{4}\right)$$

Solution
Question 8
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Two conducting sphere of radii $$ r_1$$ and $$ r_2 $$ have same electric field near their surfaces. The ratio of their electrical potentials is

A
$$ r_1^2/r^2_2 $$

B
$$ r_2^2/r^1_2 $$

C
$$ r_1/r_2 $$

D
$$ r_2/r_1 $$

Solution

$$ E_1=E_2 \Rightarrow \frac{kQ_1}{r_1^2}= \frac{kQ_2}{r_2^2} $$
$$ \frac { { Q }_{ 1 } }{ { Q }_{ 2 } } =\left( \frac { { r }_{ 1 } }{ { r }_{ 2 } } \right) ^{ 2 }$$
$$ \frac { { V }_{ 1 } }{ { V }_{ 2 } } =\frac { Q_{ 1 } }{ r_{ 1 } } \times \frac { { r }_{ 2 } }{ { Q }_{ 2 } } =\left( \frac { { r }_{ 1 } }{ { r }_{ 2 } } \right) ^{ 2 }\frac { { r }_{ 2 } }{ { r }_{ 1 } } =\frac { { r }_{ 1 } }{ { r }_{ 2 } } $$
Question 9
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Directions For Questions

$$C_1, C_2, C_3$$ and $$C_4$$ are four capacitors connected to a battery of constant e.m.f. equal to $$12$$V, as shown in Fig. Given $$C_1=1\mu F$$, $$C_2=2\mu F$$, $$C_3=3\mu F$$ and $$C_4=4\mu F$$.

...view full instructions

Find the charge on $$C_1$$ when only $$S_1$$ is closed.

A
$$16\mu C$$

B
$$9\mu C$$

C
$$6\mu C$$

D
$$8\mu C$$

Solution
Question 10
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Directions For Questions

In Fig., each capacitance $$C_1$$ is $$6.0\mu F$$, and each capacitance $$C_2$$ is $$4.0\mu F$$.

...view full instructions

The equivalent capacitance of the network between points a and b is?

A
$$2\mu F$$

B
$$4\mu F$$

C
$$6\mu F$$

D
$$8\mu F$$