Electrostatic Potential and Capacitance Test

Result

Electrostatic Potential and Capacitance Test - 50

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

A capacitor stores $$40$$ $$\mu$$C charge when connected across a battery. When the gap between the plates is filled with a dielectric a charge of $$80$$ $$\mu$$C flows through the battery. The dielectric constant of dielectric inserted is?

A
$$2$$

B
$$3$$

C
$$1$$

D
$$4$$

Solution

$$\because \displaystyle C_0=\frac{\varepsilon_0A}{d}$$ and $$C=\displaystyle\frac{\varepsilon A}{d}=\frac{K\varepsilon_0A}{d}$$
So, $$\displaystyle q_0=C_0V=\frac{\varepsilon A}{d}V$$
and $$\displaystyle q=CV=\frac{K\varepsilon_0A}{d}V$$
$$\therefore \displaystyle\frac{q}{q_0}=K$$
Given $$q_0=40\mu C$$
On inserting a dielectric, an additional charge of $$80\mu C$$ also flows through the battery.
So, total charge $$=q_0+80=40+80=120\mu C$$
$$\therefore K=\displaystyle\frac{q}{q_0}=\frac{120}{40}=3$$.
Question 2
1 / -0

The equivalent capacitance between points $$A$$ and $$B$$ will be

A
$$10\mu F$$

B
$$15\mu F$$

C
$$10.8\mu F$$

D
$$69\mu F$$

Solution

The given circuit can be redrawn as
It is a balance Wheatstone bridge type network
i.e., $$\dfrac {C_{1}}{C_{2}} = \dfrac {C_{3}}{C_{4}} = \dfrac {1}{2}$$
$$\therefore 24\mu F$$ capacitor can be neglected
Hence, equivalent capacitance between $$A$$ and $$B = 4 + 6 = 10\mu F$$.
Question 3
1 / -0

Charges $$1\mu C$$ are placed at each of the four corners of a square of side $$2\sqrt {2}m$$. The potential at the point of intersection of the diagonals is _____ $$(K = 9\times 10^{9} SI\ unit)$$.

A
$$18\times 10^{3}V$$

B
$$1800\ V$$

C
$$18\sqrt {2}\times 10^{3}V$$

D
None of these

Solution

Key Concept Here, we have to find the electric potential due to multiple of charges, so, we have to apply superposition principle to calculate net electric potential.
Length of the diagonals $$= \sqrt {2}\times$$ side length
$$= \sqrt {2} \times 2\sqrt {2} = 4\ m$$
$$\therefore$$ Length of semi-diagonals $$= \dfrac {4}{2} = 2m$$.
Now, potential due to charges at point $$O$$
$$V_{0} = \dfrac {!}{4\pi \epsilon_{0}} \left (\dfrac {1\times 10^{-6}}{2}\right ) \times 4$$
$$= 9\times 10^{9}\times 10^{-6} \times 2 = 18\times 10^{3}V$$.
Question 4
1 / -0

A hollow metal sphere of radius $$10$$cm is charged such that the potential on its surface becomes $$80$$V. The potential at the centre of the sphere is?

A
$$80$$V

B
$$800$$V

C
$$8$$V

D
Zero

Solution

The potential at the centre of the sphere is $$80$$V because it remains same at each point under the metallic hollow sphere.
Question 5
1 / -0

For the given circuit the resultant capacitance between X and Y is

A
$$ 1.6 \mu F$$

B
$$ 6.4 \mu F$$

C
$$3.2 \mu F$$

D
$$ 15 \mu F$$

Solution
Question 6
1 / -0

Find out the equivalent capacitance between $$X$$ and $$Y$$.

A
$$10/3\mu F$$

B
$$6\mu F$$

C
$$8\mu F$$

D
$$26\mu F$$

Solution

The given circuit can be redrawn as
$$C_{XY} = 8\mu F$$.
Question 7
1 / -0

Four metallic plates each with a surface area of one side $$A$$ are placed at a distance $$d$$ from each other as shown in figure. Then the capacitance of the system between $$X$$ and $$Y$$ is

A
$$\dfrac {2\epsilon_{0}A}{d}$$

B
$$\dfrac {2\epsilon_{0}A}{3d}$$

C
$$\dfrac {3\epsilon_{0}A}{d}$$

D
$$\dfrac {3\epsilon_{0}A}{2d}$$

Solution

The given circuit can be redrawn as
$$C_{eq} = \dfrac {3C}{2} = \dfrac {3\epsilon_{0}A}{2d}$$.
Question 8
1 / -0

Four conducting plates named P, Q, R and S have an area A each and are placed at the same distance d apart. The plates P and S are connected by a conducting wire and Q and R are connected to the terminals of a battery. The equivalent capacitance of the arrangement is?

A
$$\displaystyle\frac{3\varepsilon_0A}{d}$$

B
$$\displaystyle\frac{3\varepsilon_0A}{2d}$$

C
$$\displaystyle\frac{2\varepsilon_0A}{3d}$$

D
$$\displaystyle\frac{\varepsilon_0A}{3d}$$

Solution

The capacitance C of each of the three capacitors is $$\displaystyle\frac{\varepsilon_0A}{d}$$. The circuit shown in the figure can be redrawn as shown here. Now, the equivalent capacitance of this circuit is easily worked out as $$\displaystyle\frac{3C}{2}$$ or $$\displaystyle\frac{3\varepsilon_0A}{2d}$$.
Question 9
1 / -0

The following arrangement consists of five identical metal plates. Area of each plate is A and separation between the successive plates is d. The capacitance between P and

A
$$\displaystyle\frac{5\varepsilon_0A}{d}$$

B
$$\displaystyle\frac{7}{3}\varepsilon_0\frac{A}{d}$$

C
$$\displaystyle\frac{4}{3}\frac{\varepsilon_0A}{d}$$

D
$$\displaystyle\frac{5}{3}\frac{\varepsilon_0A}{d}$$

Solution

For IMAGE 02 $$C_p=C+C=2$$
FOR IMAGE 03 series combination $$C_s=\cfrac{C\times 2C}{C+2C}=\cfrac{2}{3}C$$
FOR IMAGE 04 parallel combination $$C_p=C+\cfrac{2}{3C}C=\cfrac{5}{3}C=\cfrac{5}{3}\cfrac{\varepsilon_oA}{d}$$
Question 10
1 / -0

The equivalent capacitance between points a and b in given arrangement of plates will be:(each plate has the area 'A' & separation between two adjacent plates is 'd').

A
$$\displaystyle\frac{2}{3}\frac{\varepsilon_oA}{d}$$

B
$$\displaystyle\frac{3}{2}\frac{\varepsilon_oA}{d}$$

C
$$\displaystyle\frac{2\varepsilon_oA}{d}$$

D
$$\displaystyle\frac{\varepsilon_oA}{3d}$$

Solution

FOR IMAGE 02 series combination $$C_s=\cfrac{C\times C}{2C}$$
FOR IMAGE 03 parallel combination $$C_p=C+\cfrac{C}{2}=\cfrac{3C}{2}=\cfrac{3\varepsilon_oA}{2d}$$