Electric Charges and Fields Test

Result

Electric Charges and Fields Test - 89

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Weekly Quiz Competition

Question 1
1 / -0

A proton and an electron are placed $$1.6 cm$$ apart in free space. Find the magnitude of the electrostatic force between them. The nature of this force.

A
$$ 9 \times 10^{-23} N /attraction $$

B
$$ 90 \times 10^{-23} N/attraction $$

C
$$ 9 \times 10^{-25}N/ attraction $$

D
$$ 9 \times 10^{23} N/attraction $$

Solution
Question 2
1 / -0

In a region the electric potential is given by $$V = 2x + 2y - 3z$$ obtain the expression for electric field.

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

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

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

D
None of these

Solution
Question 3
1 / -0

A long thread carrying a uniform charge $$\lambda$$ per unit length has the configuration shown in the figure. An element of charge $$q$$ is cut from the thread. The cutting portion is shown as small gap $$(AB)$$. The electric field at the point $$O$$ is

A
Zero

B
$$\dfrac {q}{4\pi \varepsilon_{0}R^{2}}$$

C
$$\dfrac {q}{2\pi \varepsilon_{0}R^{2}}$$

D
None of these

Solution
Question 4
1 / -0

In a region, an electric field $$\vec {E} = E_{0}\hat {i}$$ is present. The potential of points $$A(a, 0, 0), B(0, b, 0), C(0, 0, c)$$ and $$D(-a, 0, 0)$$ and $$V_{A}, V_{B}, V_{C}$$ and $$V_{D}$$ respectively, then

A
$$V_{A} = V_{B} = V_{C} = V_{D}$$

B
$$V_{B} = V_{C}$$

C
$$V_{D} < V_{B} < V_{A}$$

D
None of these

Solution
Question 5
1 / -0

The electric field and the electric potential at a point are E and V respectively. then the incorrect statements are

A
If $$E=0, V$$ must be zero

B
If $$V=0, E$$ must be zero

C
If $$E\neq 0, V$$ cannot be zero

D
If $$V\neq 0, E$$ cannot be zero

Solution
Question 6
1 / -0

Electric field in a plane varies like $$(2x\hat {i} + 2y\hat {j})N/C$$. If potential at infinity is taken as zero, potential at $$x = 2m, y = 2m$$ is

A
$$8\ V$$

B
$$-8\ V$$

C
Zero

D
Infinity

Solution
Question 7
1 / -0

A certain charge $$Q$$ is to between divided into two parts $$q$$ and $$Q-q$$. What is the relationship $$Q$$ and $$q$$ if the two parts, placed at a given distance $$r$$ apart, are to have the maximum Coulomb repulsion?

A
$$q=Q/2$$

B
$$q=Q/3$$

C
$$q=2Q/2$$

D
$$q=Q/4$$

Solution
Question 8
1 / -0

A point charge of $$-4.00 \,nC$$ is located at $$(0, 1.00) \,m$$.
What is the $$x$$ component of the electric field due to the point charge at $$(4.00, -2.00) \,m$$?

A
$$1.15 \,N/C$$

B
$$-0.864 \,N/C$$

C
$$1.44 \,N/C$$

D
$$-1.15 \,N/C$$

E
$$0.864 \,N/C$$

Solution

The displacement from the $$-4.00 \,nC$$ charge at point $$(0, 1 00) \,m$$ to the point $$(4.00, -2.00) \,m$$ has components $$r_x=(x_f-x_i)=+4.00 \,m$$ and $$r_y=(y_f-y_i)=-3.00 \,m$$, so the magnitude of this displacement is $$r=\sqrt{r_x^2+r_y^2}=5.00 \,m$$ and its direction is $$\theta=\tan^{-1}\left(\dfrac{r_y}{r_x}\right)=-36.9^{o}$$. The $$x$$ component of the electric field at point $$(4.00, -2.00) \,m$$ is then
$$E_x=E \cos \theta=\dfrac{k_e q}{r^2}\cos \theta$$
$$=\dfrac{(8.99\times 10^9 N.m^2/C^2)(-4.00\times 10^{-9} C)}{(5.00 \,m)^2}\cos(-36.9^{o})$$
$$=-1.15 \,N/C$$
Question 9
1 / -0

If a glass plate is placed between two charges. Then, compare the electric force with the initial condition. It will be :

A
more

B
less

C
zero

D
infinite

Solution

On placing the glass plate between the charge, the electric field will decrease as a comparison to the previous.
Question 10
1 / -0

Three charged particles are arranged on corners of a square as shown in Figure with charge $$-Q$$ on both the particle at the upper left corner and the particle at the lower right corner and with charge $$+2Q$$ on the particle at the lower left corner.
What is the direction of the electric field at the upper right corner, which is a point in empty space?

A
It is upward and to the right

B
It is straight to the right.

C
It is straight downward.

D
It is downward and to the left.

E
It is perpendicular to the plane of the picture and outward.

Solution

The charge at the upper left creates at the field point an electric field to the left, with magnitude we call $$E_1$$. The charge at lower right creates a downward electric field with an equal magnitude $$E_1$$. These two charges together create a field $$\sqrt{2}E_1$$ downward and to the left $$(at \,45^{o})$$. The positive charge has twice the charge but is $$\sqrt{2}$$ time farther from the field point, so it creates a field $$2E_1/(\sqrt{2})^2=E_1$$ upward and to the right. The fields from the two charges are opposite in direction, and the field from the negative charges is stronger, so the net field is then $$(\sqrt{2}-1)E_1$$, which is downward and to the left $$(at \,45^{o})$$.