Electric Charges and Fields Test

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Electric Charges and Fields Test - 37

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Question 1
1 / -0

$$5$$ charges each of magnitude $$10^{-5}$$ $$C$$ and mass $$1 kg$$ are placed (fixed) symmetrically about a movable central charges of magnitude $$5 \times 10^{-5}C$$ and mass $$0.5 kg$$ as shown. The charge at $$P_{1}$$, is removed. The acceleration of the central charge is

$$[$$Given $$OP_1 = OP_2 = OP_3 = OP_4 = OP_5 = 1m;$$$$\displaystyle \frac{1}{4\pi \varepsilon _{0}}=9\times 10^{9}$$ in $$SI$$ units$$]$$

A
$$9 m s^{-2}$$ upwards

B
$$9 m s^{-2}$$ downwards

C
$$4.5 m s^{-2}$$ upwards

D
$$4.5 m s^{-2}$$ downwards

Solution

Now Force on Q = Force in figure (1) + Force in fugure (2)
F = $$0\quad +\quad \frac { KqQ }{ { R }^{ 2 } } \quad \uparrow upward\\$$
F = $$0\quad +\quad \frac { \left( 9\times { 10 }^{ 9 } \right) \times { 10 }^{ -5 }\times \left( 5\times { 10 }^{ -5 } \right) }{ 1 } \\$$
$$ma\quad =\quad 9\times { 10 }^{ -1 }\times 5\\ 0.5a\quad =\quad 9\times { 10 }^{ -1 }\times 5\\ a\quad =\quad 9\frac { m }{ { sec }^{ 2 } } \quad \uparrow upward$$
Question 2
1 / -0

Two particles having charges $$q_{1}$$ and $$q_2$$ when kept at a certain distance, exert force $$F$$ on each other. If distance is reduced to half, force between them becomes :

A
$$\displaystyle \frac {F}{2}$$

B
$$2F$$

C
$$4F$$

D
$$\displaystyle \dfrac {F}{4}$$

Solution

Answer is C.
Coulomb's law describes the magnitude of the electrostatic force between two electric charges. The Coulomb's law formula is given as follows.

The force between $$2$$ charges $$q_1$$ and $$q_2$$ is given by the formula $$F=k\dfrac { q_1\times q_2 }{ { r }^{ 2 } } $$.

Where, $$q_1$$: Charge of object $$1$$

$$q_2$$: Charge of object $$2$$

r: Distance between the two objects

F: Force between the two objects. A positive force implies a repulsive interaction, while a negative force implies an attractive interaction

k: Coulomb Constant.

In this case, the distance between $$2$$ charges is halved. So the new force becomes $${ F }_{ new }=k\dfrac { q_1\times q_2 }{ { (\dfrac { r }{ 2 } ) }^{ 2 } } \quad =\quad \dfrac { 4\times k\times q_1\times q_2 }{ { r }^{ 2 } } \quad =\quad 4F$$.

Hence, the force between the charges becomes $$4F$$.
Question 3
1 / -0

Two charges of $$+1$$ $$\mu C$$ $$\&+5$$ $$\mu C$$ are placed $$4 cm$$ apart, the ratio of the force exerted by both charges on each other will be -

A
$$1:1$$

B
$$1:5$$

C
$$5:1$$

D
$$25:1$$

Solution

From Newton's Third law of motion , force exerted by charge $$1\mu C$$ on
charge $$5\mu C$$ is equal to the force by charge $$5\mu C$$ on charge
$$1\mu c$$ in magnitude.

Hence ratio of force is $$1:1$$

Answer-(A)
Question 4
1 / -0

Two identical charges of magnitude +Q are fixed as shown . A third charge -Q is placed mid way between them at point P. Then small displacements of -Q are made in the directions indicated by arrows. The -Q is stable with respect to displacement .

A
I and III

B
I and V

C
II and $$IV$$

D
III and V

E
stable for any small displacement.

Solution

If we displace through a displacement d then a restoring force acts to make -Q to come back to make it stable at their.
Same would happen if it goes downward.
Question 5
1 / -0

If two non-identical charges are kept at a certain distance $$d$$, which one experiences more force?

A
Smaller charge

B
Larger charge

C
Both charges experience same force

D
Cannot determine

Solution

By Coulomb's law, the force on $$q_1$$ due to $$q_2$$ is $$F_1=q_1E_2=q_1\dfrac{kq_2}{r^2}=\dfrac{kq_1q_2}{r^2}$$ where $$E_1$$ be the electric filed at $$q_1$$ due to $$q_2$$

similarly, force on q_2 is $$F_2=q_2E_1= q_2\dfrac{kq_1}{r^2}=\dfrac{kq_1q_2}{r^2}$$ where $$E_2$$ be the electric filed at $$q_2$$ due to $$q_1$$

Thus, $$F_1=F_2$$
Question 6
1 / -0

When the distance between two charged particles is halved, then the force between them becomes :

A
One-fourth

B
Half

C
Double

D
Four times

Solution

Earlier force between two particles is

$$F= \dfrac{kq_1q_2}{r^2}$$

When distance is halved force becomes

$$F'=\dfrac{kq_1q_2}{(r/2)^2}=4 \dfrac{kq_1q_2}{r^2}=4 F$$
Question 7
1 / -0

Two charges are placed a certain distance apart in air. Ifa glass slab is introduced between them, the force between them will

A
increase

B
decrease

C
remains the same

D
become zero

Solution

In air , the force between charges is $$F=\dfrac{q_1q_2}{4\pi\epsilon_0 r^2}$$
When the glass slab is introduced between them , the permeability becomes $$\epsilon=K\epsilon_0$$ where $$K=$$ dielectric constant of glass.
Thus, force becomes, $$F'=\dfrac{q_1q_2}{4\pi K\epsilon_0 r^2}$$
Hence, $$F'<F$$
Question 8
1 / -0

$$12$$ positive charges of magnitude $$q$$ are placed on a circle of radius $$R$$ in a manner that they are equally spaced. $$A$$ charge $$+Q$$ is placed at the centre. If one of the charges $$q$$ is removed, then the force on $$Q$$ is :

A
Zero

B
$$\displaystyle \frac {qQ}{4\pi \varepsilon _{0}R^{2}}$$ away from the position of the removed charge.

C
$$\displaystyle \frac {11qQ}{4\pi \varepsilon _{0}R^{2}}$$ away from the position of the removed charge.

D
$$\displaystyle \frac {qQ}{4\pi \varepsilon _{0}R^{2}}$$ towards the position of the removed charge.

Solution

Now force on charge Q = force in figure (i) + force in figure (ii)
F = $$0\quad +\quad \frac { KqQ }{ { R }^{ 2 } }$$
F = $$ \frac { KqQ }{ { R }^{ 2 } }$$ towards the position of that removed charge.
Question 9
1 / -0

Two charges of 10C and -15C are separated in air by 1 m. The ratio of magnitude of force exerted by one on the other is

A
1 : 2

B
2 : 1

C
1 : 1

D
none of these

Solution

The electric forces on between two charges due to each other is same and opposite. i,e $$|F_{12}|=|-F_{21}| $$
Thus the ratio will be $$1:1$$
Question 10
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In the given figure calculate the force on unit positive charge placed at $$A$$. The length of each side on the square is $$4 cm$$.

A
$$\displaystyle 9\sqrt{3} \times 10^{4} dyn$$

B
$$\displaystyle 9\sqrt{3} \times 10^{-4} dyn$$

C
$$\displaystyle 9\sqrt{3} \times 10^{4} N$$

D
$$\displaystyle 9\sqrt{3} \times 10^{-4} N$$

Solution