Electric Charges and Fields Test

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Electric Charges and Fields Test - 79

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Question 1
1 / -0

If the electric potential in a region is represented as $$V = 2x 3y 4z$$, then the electric field strength is:

A
$$-2\hat{i}+3\hat{j}-4\hat{k}$$

B
$$-2\hat{i}-3\hat{j}-4\hat{k}$$

C
$$-2\hat{i}-3\hat{j}+4\hat{k}$$

D
$$2\hat{i}+3\hat{j}-4\hat{k}$$

Solution
Question 2
1 / -0

A force of 4 N is active between two charges in air.If the space between them is completely filled with glass $$\left( { \varepsilon }_{ r }=8 \right) $$, then the new force will be

A
2 N

B
5 N

C
0.2 N

D
0.05 N

Solution
Question 3
1 / -0

Consider a hollow sphere of radius $$R$$, Its quarter part is cut and kept that geometrical centre lies at origin as shown in figure.
$$x-y$$ axis lies in plane of paper and $$z$$-axis is $$\bot $$ to the plane of the plane the paper. Sphere is kept in a manner such that plane portions of sphere lies in $$x-z$$ and $$y-z$$ plane. Uniform electric field $$\vec{E}=E_{0}\hat{i}+2E_{0}\hat{j}+2E_{0}\hat{k}$$ $$(N/c)$$ lies in the complete space Electric flux coming out of the curved surface of sphere is:

A
$$\dfrac{3}{2}\pi R^{2}E_{0}$$

B
$$\dfrac{5}{2}\pi R^{2}E_{0}$$

C
$$2\pi R^{2}E_{0}$$

D
$$Zero$$

Solution

Magnitude of electric flux entering the XZ plane surface and YZ plane will be equal to electric flux coming out of curved surface.
Flux through XZ plane,
Area vector for XZ plane, $$ A = -\dfrac{\pi R^2}{2} \hat{j}$$

$$\phi_1 = E.A = (E_0\hat{i} + 2E_0\hat{j} +2E_0\hat{k}).(-\dfrac{\pi R^2}{2}\hat{j})$$

$$\phi_1 = -\pi R^2E_0$$

Flux through YZ plane,
Area vector for YZ plane, $$ A = -\dfrac{\pi R^2}{2} \hat{i}$$

$$\phi_2 = E.A = (E_0\hat{i} + 2E_0\hat{j} +2E_0\hat{k}).(-\dfrac{\pi R^2}{2}\hat{i})$$

$$\phi_2 = -\dfrac{\pi R^2E_0}{2}$$

Total flux, $$\phi = \phi_1+\phi_2 = \dfrac{3\pi R^2E_0}{2}$$
Question 4
1 / -0

A charge Q is placed at the centre of the open end of a cylindrical vessel of radius R and height 2R as shown in figure. the flux of the electric field through the surface ( curved surface + base ) of the vessel is

A
$$ \dfrac { Q }{ \varepsilon_0 } $$

B
$$ \dfrac { Q }{ 2 \varepsilon_0 } $$

C
$$ \dfrac { Q }{ 4 \varepsilon_0 } $$

D
$$ \dfrac { Q }{ \sqrt {5} \varepsilon_0 } $$

Solution

In the figure, we have shown in the figure part a charge $$q$$ at the centre of one open end of a cylinder. Note that the other end is closed.
Now, take an another identical cylinder and joint's open end with the open of end of the first end of the first cylinder. Now, we have closed surface of two cylinders.
This electric flux through this closed surface, according to Gauss's according to Gauss's theorem is $$q/{ \epsilon }_{ 0 }$$ where $${ \epsilon }_{ 0 }$$ is electric permittivity of space.
Then, from the symmetry, we can say that flux through surface of any one of the two joined surface will be $$\dfrac { q }{ 2{ \epsilon }_{ 0 } } $$.
Question 5
1 / -0

Consider the situation as shown in the figure. If the current $$I$$ in long straight conducting wire XY is increased at a steady rate then the induced e.m.f.'s in loops A and B will be

A
clockwise in A, anti clockwise in B

B
anti clockwise in A, clockwise in B

C
Both A and B

D
None of these

Solution
Question 6
1 / -0

If the field are parallel to surface then electric flux, linked linked, with the surface will be

A
Greater than zero

B
Equal to zero

C
May be less than zero

D
Can't determine

Solution
Question 7
1 / -0

Two point charges $$q_{1}$$ and $$q_{2}$$ are separated by a distance $$r$$. The electric force on $$ q_{1}$$ by $$ q_{2}$$ is $$\overrightarrow{F_{1}}$$ and on $$ q_{2}$$ by $$ q_{1}$$ is $$\overrightarrow{F_{2}}$$. Then

A
$$\overrightarrow{F_{1}}=\overrightarrow{F_{2}}$$

B
$$\overrightarrow{F_{1}}.\overrightarrow{F_{2}}>0$$

C
$$\overrightarrow{F_{1}}.\overrightarrow{F_{2}}<0$$

D
$$\overrightarrow{F_{1}}\times\overrightarrow{F_{2}}<0$$

Solution
Question 8
1 / -0

A cylinder of length $$l$$, radius $$R$$ is kept in the uniform electric field as shown in the figure. If electric field strength is $$E$$, then the outgoing electric flux through the cylinder is

A
$$Zero$$

B
$$\pi R^{2}E(l)$$

C
$$R/ \pi$$

D
$$2R/E$$

Solution
Question 9
1 / -0

What will be the total flux through faces of the cube as given in the figure with side of length $$'a'$$ if charge $$q$$ is placed at a corner of the cube

A
$$\frac { q } { 8 \varepsilon _ { 0 } }$$

B
$$\frac { q } { 4 \varepsilon _ { 0 } }$$

C
$$\frac { q } { 2 \varepsilon _ { 0 } }$$

D
$$\frac { q } { 3 \varepsilon _ { 0 } }$$

Solution
Question 10
1 / -0

two large, thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite signs and of magnitude $$16\times{10}^{22}{cm}^{2}$$. The electric field between the plates is .

A
$$1.8\times{10}^{-10}{NC}^{-1}$$

B
$$1.9\times{10}^{-10}{NC}^{-1}$$

C
$$1.6\times{10}^{-10}{NC}^{-1}$$

D
$$1.5\times{10}^{-10}{NC}^{-1}$$

Solution