Electrostatic Potential and Capacitance Test

Result

Electrostatic Potential and Capacitance Test - 61

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Weekly Quiz Competition

Question 1
1 / -0

The electric potential decreases uniformly from 120 V to 80 V as one moves on the X-axis from $$ x = -1 cm$$ to $$x = +1$$ cm. The electric field at the origin.

A
must be equal to $$20 V/cm$$

B
must be equal to $$2.0 V/cm$$

C
must be greater than $$20 V/cm$$

D
must be less than $$20 V/cm$$

Solution

E = -dV/dx

Since V decreases uniformly, E = -ΔV/Δx

= -(80 - 120)/{1- (-1)}

= - ( -40)/2
= 20 V/Cm in the positive direction of x axis
Question 2
1 / -0

The equivalent capacitance between A and B in the circuit given below, is:

A
$$2.4$$ $$\mu F$$

B
$$4.9$$ $$\mu F$$

C
$$3.6$$ $$\mu F$$

D
$$5.4$$ $$\mu F$$

Solution
Question 3
1 / -0

Let $$V_0$$ be the potential at the origin in an electric field $$\overset{\rightarrow}{E}=E_x\hat{i}+E_y\hat{j}$$. The potential at the point $$(x,y)$$ is:

A
$$V_0-{_{x}{E}_x}-{_{y}{E}_y}$$

B
$$V_0+{_{x}{E}_x}+{_{y}{E}_y}$$

C
$${_{x}{E}_x}+{_{y}{E}_y}-V_0$$

D
$$\left(\sqrt{x^2+y^2}\right)\sqrt{{E}_{x}^{2}+{E}_{y}^{2}}-V_0$$

Solution
Question 4
1 / -0

The electric potential at a distance of $$3 \,m$$ on the axis of a short dipole of dipole moment $$4\times 10^{-12}$$ coulomb-metre is

A
$$1.33 \times 10^{-3} V$$

B
$$4 \,mV$$

C
$$12 \,mV$$

D
$$27 \,mV$$

Solution

$$d=3m$$
dipole moment $$=4\times { 10 }^{ -12 }cm$$
$$Q=\dfrac { P }{ d } =\dfrac { 4\times { 10 }^{ -12 }cm }{ 3m } $$
$$=1.33\times { 10 }^{ -12 }C$$
Question 5
1 / -0

Two charges $$-5\mu C$$ and $$+10\mu C$$ are placed 20 cm apart. The net electric field at the mid-point between the two charge is

A
$$4.5\times { 10 }^{ 5 }N/C$$ directed towards $$+10\mu C$$

B
$$13.5\times { 10 }^{ 5 }N/C$$ directed towards $$-5\mu C$$

C
$$13.5\times { 10 }^{ 5 }N/C$$ directed towards $$+10\mu C$$

D
$$4.5\times { 10 }^{ 5 }N/C$$ directed towards $$-5\mu C$$

Solution

R.E.F image
$$ \bar{E}net = \bar{E_{1}} + \bar{E_{2}} $$
$$ = \frac{K(5 \times 10^{-6})}{10 \times 10^{-2}} + \frac{K(10 \times 10^{-6})}{10 \times 10^{-2}}$$
$$ \Rightarrow 9\times 10^{9} [5 \times 10^{-5} + 10 \times 10^{-5}]$$
$$ \Rightarrow 135 \times 10^{4} $$ N\C
$$ \bar{E}net $$ $$ 13.5 \times $$ $$ 10^{5} $$ N/C towards $$ -5\mu c$$
Question 6
1 / -0

In the following question the value of equivalent capacitane between point a & b is:-

A
$$30 \mu F$$

B
$$6 \mu F$$

C
$$18 \mu F$$

D
$$12 \mu F$$
Question 7
1 / -0

In the circuit diagram given below, the value of the potential difference across the plates of the capacitors are

A
17.5 KV, 7.5 KV

B
10 KV, 15 KV

C
5 KV, 20 KV

D
16.5 KV, 8.5 KV

Solution
Question 8
1 / -0

A parallel plate capacitor of area $$60 cm^2$$ and separation 3 mm is charged initially to $$90 \mu C$$. If medium between the plates gets slightly conducting and the p!ate loses the charge initially at rate of $$2.5\times10^{-8} C/s$$, then what is the magnetic field between the plates?

A
$$2.5\times10^{-8} T$$

B
$$2.0\times10^{-7} T$$

C
$$1.63\times10^{-11} T$$

D
ZERO

Solution

If the charge starts flowing its kind of current carrying wire so they won't be any magnetic field in the wire(plate) but outside is the magnetic field. magnetic field between two plates is zero.
Question 9
1 / -0

Then output in the circuit of figure is taken across a capacitor. It is as shown in figure:-

A

B

C

D

Solution

As $$RC$$ time constant of the capacitor is quite large $$( \tau =RC =10\times 10^3 \times 10 \times 10^{-6}=0.1\ sec)$$, if will not discharge appreciably. Hence voltage remains nearly constant and option $$C$$ is correct.
Question 10
1 / -0

Calculate the equivalent capacity of the system of capacitors shown across A and B

A
C

B
2C

C
3C

D
None of these

Solution