Electric Charges and Fields Test

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Electric Charges and Fields Test - 42

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Weekly Quiz Competition

Question 1
1 / -0

The Electrostatic Force between two charges is 4 N. If one charge has a value of 5 $$\mu$$C and is 2 cm from the second charge, find the new electrostatic force if the second charge is moved 2 cm further away from the first charge.

A
1 N

B
2 N

C
4 N

D
8 N

E
16 N

Solution

By Coulomb's law, $$F_1=\dfrac{kq_1q_2}{r^2}=4$$
In the second case, $$r=2r$$, so the force will be $$F_2=\dfrac{kq_1q_2}{(2r)^2}=F_1/4=4/4=1 N$$
Question 2
1 / -0

A charge $$\mathit{q}$$ of mass $$\mathit{m}$$ that is free to move is placed at the origin of a coordinate system.
A second identical charge, but of twice the mass, is placed at a distance $$d$$ along the positive $$x$$ axis from the origin.
Which of the following correctly describes the magnitude and direction of the force acting on the particle at the origin?

A
$$\mathit{\cfrac{2kq}{d^2}}$$, to the left

B
$$\mathit{\cfrac{2kq^2}{d^2}}$$, to the right

C
$$\mathit{\cfrac{2kq^2}{d^2}}$$, to the left

D
$$\mathit{\cfrac{kq^2}{d^2}}$$, to the right

E
$$\mathit{\cfrac{kq^2}{d^2}}$$, to the left

Solution

We know that the electric force $$F$$ between two charges$$q_{1} , q_{2}$$ separated by a distance $$r$$ is given by ,
$$F=kq_{1}q_{2}/r^{2}$$ ,
given that $$q_{1}=q , q_{2}=q , r=d$$ ,
therefore force between them is
$$F=kq\times q/d^{2}$$ ,
or $$F=kq^{2}/d^{2}$$ ,
this will be the force on the particle at origin , now as the two charges are positive therefore there is a repulsion force between them hence force on q at origin will be towards the left .
Question 3
1 / -0

The Electrostatic Force between two identical charges $$q_0$$ separated by a distance $$r_0$$ is $$F_0$$. The magnitudes of each charge and the separation distance are tripled. Find the new Electrostatic Force in terms of $$F_0$$.

A
$$\cfrac{1}{9}F_0$$

B
$$\cfrac{1}{3}F_0$$

C
$$F_0$$

D
$$3F_0$$

E
$$9F_0$$

Solution

Let the distance between the charges initially be $$r$$.
Thus initial electrostatic force $$F_o = \dfrac{kQ_1Q_2}{r^2}$$

New distance between the charges $$r'= 3r$$
Charge of each particle is tripled $$Q'_1 = 3Q_1$$ and $$Q_2' =3Q_2$$

$$\therefore$$ New electrostatic force $$F' = \dfrac{k(3Q_1)(3Q_2)}{(3r)^2} = F_o$$
Question 4
1 / -0

Coaxial cable (typically used for cable and satellite tv) has its signal run on a copper wire surrounded by an insulator which is surrounded by the ground wire, as opposed to the typical side by side configuration.
What is the most logical reason for this?

A
The energy held in the electric field between the inside wire and the outside allow for a clearer signal.

B
The outside wire prevents any magnetic field from the inside wire from leaking out of the wire.

C
The outside wire prevents electric fields from interfering with the signal in the inside wire.

D
The overall resistance of the wire is reduced using this configuration.

Solution

Coaxial Cable is type of electric cable that has an inner conductor surrounded by a tubular insulating layer, surrounded by a tubular conducting shield. The energy held in the electric field between the inside wire and outside allow for a clearer signal.
Question 5
1 / -0

Compare how the magnitude of the electric field decreases with distance from a uniform charged sphere and a uniform charged wire.

A
Both decrease with a one over the square of the distance relationship.

B
From the sphere, the electric field decreases with a one over the distance relationship. From the wire, the electric field decreases with a one over the distance squared relationship.

C
Both decrease with a one over the distance relationship.

D
From the wire, the electric field decreases with a one over the distance relationship. From the sphere, the electric field decreases with a one over the distance squared relationship.

Solution

Electric field at a point due to uniform charged wire $$E_w = \dfrac{\lambda}{2\pi \epsilon_o r}$$ $$\implies E_w =\propto \dfrac{1}{r}$$
Thus electric field decreases with one over the distance relationship.

Electric field at a point due to uniform charged sphere $$E_s = \dfrac{Q}{4\pi \epsilon_o r^2}$$ $$\implies E_s= \propto \dfrac{1}{r^2}$$
Thus electric field decreases with one over the distance squared relationship.
Question 6
1 / -0

Two identical conducting spheres were situated so that they were insulated from their surroundings, and their centers were separated by a distance, d.
One sphere had a charge of $$+8$$ Coulombs; the other had a charge of $$-4$$ Coulombs.
The spheres were then brought together to touch each other. The spheres were separated and again insulated from their surroundings, and their centers were separated by a distance d.
How does the magnitude of the electrostatic force between the spheres in their final situation compare to the magnitude of electrostatic force between them in their original situation?

A
The magnitude of the final force is the same as the magnitude of the original force

B
The magnitude of the final force is half as much as the magnitude of the original force

C
The magnitude of the final force is twice as much as the magnitude of the original force

D
The magnitude of the final force is one-fourth as much as the magnitude of the original force

E
The magnitude of the final force is one-eighth as much as the magnitude of the original force

Solution

Before touch , the electrostatic force between two given identical spheres ,
$$F_{before}=k8\times4/d^{2}$$ ...............................eq1
net charge of spheres $$=+8-4=+4C$$
now as the spheres are identical therefore their capacitances will be same , we know charges on spheres after touch are given by ,
$$q_{1}'/q_{2}'=C_{1}/C_{2}=1$$ (because $$C_{1}=C_{2}$$)
or $$q_{1}=q_{2}$$
it means net charge will be divided equally on both he spheres i.e. $$q_{1}=q_{2}=4/2=2C$$
therefore after touch , the electrostatic force between two spheres ,
$$F_{afer}=k2\times2/d^{2}$$ ...............................eq2
dividing eq2 by eq1 ,
$$F_{after}/F_{before}=4/32=1/8$$
or $$F_{after}=1/8F_{before}$$
Question 7
1 / -0

A 1 C charge and a 8C charge experience a force of 180 N.
How far apart are the charges?

A
20 km

B
0.02 m

C
0.2 m

D
2 m

E
4 m

Solution

Given : $$Q_1 = 1$$ C $$Q_2 = 8$$ C $$F = 180$$ N
The electrostatic force between the charges $$F = \dfrac{kQ_1Q_2}{r^2}$$ where $$k =9\times 10^9$$
$$\therefore$$ $$180 = \dfrac{9\times 10^9 \times 1\times 8}{r^2}$$ $$\implies r =20,000$$ m $$ = 20$$ km
Question 8
1 / -0

If the distance between two unlike poles is doubled then the force of attraction between them becomes ______ times its original value.

A
$$0.25$$

B
$$0.5$$

C
$$0.125$$

D
$$2$$

Solution

Force between two charges at a distance r is
$$F=\cfrac{1}{4\pi {\epsilon}_0}\cfrac{q_1q_2}{r^2}$$
when r' = 2r
then
$$F'=\cfrac{1}{4\pi {\epsilon}_0}\cfrac{q_1q_2}{{r'}^2}$$
$$F'=\cfrac{1}{4\pi {\epsilon}_0}\cfrac{q_1q_2}{4r^2}$$
$$F'=\cfrac{1}{4}F \; i.e. \; 0.25F$$
Question 9
1 / -0

______ is that charge which repels an equal and similar charge with a force of $$9\times 10^{9}N$$, when placed in vacuum at a distance of $$1\ m$$ from it.

A
One coulomb

B
Two coulomb

C
Three coulomb

D
Four coulomb

Solution

we know that force is given by $$F=\dfrac{Q_1Q_2}{4\pi\epsilon_o R^2}$$

Now, here $$Q_1=Q_2=Q$$ and

$$R=1m $$ and $$F=9\times 10^9N$$

And, $$\dfrac{1}{4\pi\epsilon_o}=9\times 10^9$$

Hence from force equation-

$$\implies \dfrac{9\times 10^9\times Q^2}{1^2}=9\times 10^9$$

$$\implies Q^2=1$$

$$\implies Q=\pm1C$$

Hence charge is $$1\hspace{2mm}coulomb$$.

Answer-(A)
Question 10
1 / -0

Two point charges separated by a distance d repel each other with a force of $$10\ N$$. If the separation between them becomes $$d/2$$, the force of repulsion will be ______.

A
$$10\ N$$

B
$$30\ N$$

C
$$40\ N$$

D
$$5\ N$$

Solution

$$F=\dfrac{q_1q_2}{4\pi\epsilon_o d^2}=10N$$

Now when distance becomes $$\dfrac{d}{2}$$-

$$F'=\dfrac{q_1q_2}{4\pi\epsilon_o \left(\dfrac{d}{2}\right)^2}$$

$$\implies F'=\dfrac{4q_1q_2}{4\pi\epsilon_o d^2}=4F=4\times 10$$

$$\implies F'=40N$$

Answer-(C)

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Electric Charges and Fields Test - 42

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Electric Charges and Fields Test - 42

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Question 1

1 N

2 N

4 N

8 N

16 N

Solution

Question 2

$$\mathit{\cfrac{2kq}{d^2}}$$, to the left

$$\mathit{\cfrac{2kq^2}{d^2}}$$, to the right

$$\mathit{\cfrac{2kq^2}{d^2}}$$, to the left

$$\mathit{\cfrac{kq^2}{d^2}}$$, to the right

$$\mathit{\cfrac{kq^2}{d^2}}$$, to the left

Solution

Question 3

$$\cfrac{1}{9}F_0$$

$$\cfrac{1}{3}F_0$$

$$F_0$$

$$3F_0$$

$$9F_0$$

Solution

Question 4

The energy held in the electric field between the inside wire and the outside allow for a clearer signal.

The outside wire prevents any magnetic field from the inside wire from leaking out of the wire.

The outside wire prevents electric fields from interfering with the signal in the inside wire.

The overall resistance of the wire is reduced using this configuration.

Solution

Question 5

Both decrease with a one over the square of the distance relationship.

From the sphere, the electric field decreases with a one over the distance relationship. From the wire, the electric field decreases with a one over the distance squared relationship.

Both decrease with a one over the distance relationship.

From the wire, the electric field decreases with a one over the distance relationship. From the sphere, the electric field decreases with a one over the distance squared relationship.

Solution

Question 6

The magnitude of the final force is the same as the magnitude of the original force

The magnitude of the final force is half as much as the magnitude of the original force

The magnitude of the final force is twice as much as the magnitude of the original force

The magnitude of the final force is one-fourth as much as the magnitude of the original force

The magnitude of the final force is one-eighth as much as the magnitude of the original force

Solution

Question 7

20 km

0.02 m

0.2 m

2 m

4 m

Solution

Question 8

$$0.25$$

$$0.5$$

$$0.125$$

$$2$$

Solution

Question 9

One coulomb

Two coulomb

Three coulomb

Four coulomb

Solution

Question 10

$$10\ N$$

$$30\ N$$

$$40\ N$$

$$5\ N$$

Solution

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