Electric Charges and Fields Test

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Electric Charges and Fields Test - 66

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Question 1
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Find the force experienced by a semicircular rod having a charge q as shown in figure. Radius of the wire is R, and the line of charge with linear charge density $$\lambda$$ passes through its center and is perpendicular to the plane of wire.

A
$$\displaystyle \frac{\lambda q}{2 \pi^2 \varepsilon_0 R}$$

B
$$\displaystyle \frac{\lambda q}{\pi^2 \varepsilon_0 R}$$

C
$$\displaystyle \frac{\lambda q}{4 \pi^2 \varepsilon_0 R}$$

D
$$\displaystyle \frac{\lambda q}{8 \pi^2 \varepsilon_0 R}$$

Solution

$$\displaystyle F_{net} = \int dq E cos \theta$$
$$= \displaystyle \int_{- \pi/2}^{\pi/2}\left ( \frac{q}{\pi R} \right ) Rd \theta \frac{\lambda}{2 \pi \varepsilon \cdot R} cos \theta$$
$$= \displaystyle \frac{\lambda q}{2 \pi^2 \varepsilon_0 R} \int_{- \pi/2}^{\pi/2} cos \theta d \theta = \frac{\lambda q}{2 \pi^2 \varepsilon_0 R} [sin \theta]_{- \pi/2}^{\pi/2}$$
$$= \displaystyle \frac{\lambda q}{2 \pi^2 \varepsilon_0 R} [1 - (-1)] = \frac{\lambda q}{\pi^2 \varepsilon_0 R}$$
Question 2
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One-fourth of a sphere of radius R is removed as shown in Fig. An electric field E exists parallel to the xy plane. Find the flux through the remaining curved part :

A
$$\pi R^2E$$

B
$$\sqrt2 \pi R^2E$$

C
$$\pi R^2E/ \sqrt2$$

D
none of these

Solution

$$\phi _{plain}+\phi _{curve}=0$$ or $$\phi _{plain}=-\phi _{curve}$$
$$\vec{A}_1=-\dfrac{\pi R^2}{2}\hat{i},\vec{A}_2=-\dfrac{\pi R^2}{2}\hat{j}$$
$$\vec{E}=E cos 45^{\circ}\hat{i}+E sin 45 ^{\circ}\hat{j}$$
$$\dfrac{E}{\sqrt 2}\hat{i}+\dfrac{E}{\sqrt 2}\hat{j}$$
$$\phi =\vec{E}\cdot (\vec{A}_1+\vec{A}_2)$$
$$\dfrac{-E}{\sqrt 2} \dfrac{\pi R^2}{2}-\dfrac{E}{\sqrt 2}\dfrac{\pi R^2}{2}=\dfrac{-\pi R^2 E}{\sqrt 2}$$
Thus is the flux entering. So flux is $$\dfrac{\pi R^2 E}{\sqrt 2}$$
Question 3
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Two identical particles are charged and held at a distance of $$1 m$$ from each other. They are found to be attracting each other with a force of $$0.027 N$$. Now, they are connected by a conducting wire so that charge flows between them. When the charge flow stops, they are found to be repelling each other with a force of $$0.009 N$$. Find the initial charge on each particle :

A

$$q_1 = \pm3 \mu C \ ; \ q_2= \mp1 \mu C$$

B

$$q_1=\mp 3 \mu C \ ; \ q_2=\mp 2 \mu C$$

C
$$q_1=\pm 5 \mu C \ ; \ q_2=\mp 2 \mu C$$

D
None of these

Solution

Let the initial charges are $$q_1$$ and $$q_2$$.
In first case , $$F=\dfrac{kq_1q_2}{r^2}=-0.027 $$ (minus sign due to attraction force)
or $$\dfrac{(9\times 10^9)q_1q_2}{1^2}=-0.027 \Rightarrow q_1q_2=-3\times 10^{-12} . . (1)$$
When a conducting wire are connected them, the charge on each particle will be half of the total charge i.e $$(q_1+q_2)/2$$.
For repulsion, $$F=\dfrac{(9\times 10^9(\dfrac{q_1+q_2}{2})^2}{1^2}=0.009$$
or $$(q_1+q_2)^2=4\times 10^{-12}$$
or $$q_1+q_2=\pm 2\times 10^{-6} ...(2)$$
A) From $$q_1q_2=-3\times 10^{-12}$$ and $$q_1+q_2=+ 2\times 10^{-6} $$, we get after solving $$q_1=3 \mu C$$ and $$q_2=-1 \mu C$$
B) From $$q_1q_2=-3\times 10^{-12}$$ and $$q_1+q_2=- 2\times 10^{-6} $$, we get after solving $$q_1=-3 \mu C$$ and $$q_2=1 \mu C$$
Question 4
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Two similarly and equally charged identical metal spheres A and B repel each other with a force of $$2 \times 10^{-5} N$$. A third identical uncharged sphere C is touched with A and then placed at the midpoint between A and B. Find the net electric force on C :

A
$$4 \times 10^{-5} N$$ away from sphere B

B
$$4 \times 10^{-5} N$$ toward sphere B

C
$$2 \times 10^{-5} N$$ away from sphere B

D
$$2 \times 10^{-5} N$$ toward sphere B

Solution

Let the charge on each sphere be q then the force on each sphere is $$F=\dfrac{kqq}{r^2}=\dfrac{kq^2}{r^2}=2\times 10^{-5} .....(1)$$
When sphere C is touched by sphere A, charge q gets equally divided between the two.
Thus, $$F_{CA}=\dfrac{k(q/2)(q/2)}{r^2/4}=\dfrac{kq^2}{r^2}$$
and $$F_{CB}=\dfrac{k(q/2)(q)}{r^2/4}=\dfrac{2kq^2}{r^2}$$
Net force $$=F_{CB}-F_{CA}=\dfrac{2kq^2}{r^2}-\dfrac{kq^2}{r^2}=\dfrac{kq^2}{r^2}=2\times 10^{-5} N $$ (using (1))
As $$F_{CB}>F_{CA}$$ so net force is away sphere B.
Question 5
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Find the electric field vector at P(a, a, a) due to three infinitely long lines of charges along the x-, y- and z-axes, respectively. The charge density, i.e., charge per unit length of each wire is $$\lambda$$ :

A
$$\displaystyle \frac{\lambda}{3 \pi \varepsilon_0 a} (\widehat i + \widehat j + \widehat k)$$

B
$$\displaystyle \frac{\lambda}{2 \pi \varepsilon_0 a} (\widehat i + \widehat j + \widehat k)$$

C
$$\displaystyle \frac{\lambda}{2 \sqrt{2} \pi \varepsilon_0 a} (\widehat i + \widehat j + \widehat k)$$

D
$$\displaystyle \frac{\sqrt{2} \lambda}{\pi \varepsilon_0 a} (\widehat i + \widehat j + \widehat k)$$

Solution

Electric field due to an infinitely long wire at a distance r is given as $$\dfrac{\lambda}{2\pi \epsilon_o r}$$
Let us consider the electric field due to the wire along the z axis at point p.
$$\vec E_z = E \widehat r$$
$$\displaystyle = \frac{\lambda}{2 \pi \varepsilon_0 (x^2 + y^2)^{1/2}} \widehat r=\frac{\lambda}{2 \pi \varepsilon_0 (a^2 + a^2)^{1/2}} (\widehat i cos 45^o + \widehat j cos 45^o)$$
$$= \displaystyle \frac{\lambda}{2 \sqrt{2} \pi \varepsilon_0 a} \frac{1}{\sqrt{2}} (\widehat i + \widehat j)$$
$$= \displaystyle \frac{\lambda}{4 \pi \varepsilon_0 a} (\widehat i + \widehat j)$$
Similarly, electric field due to wires along x and y axes is given as
$$\displaystyle

\vec E_{y} = \frac{\lambda}{4 \pi \varepsilon_0 a} (\widehat i +

\widehat k)$$ and $$\displaystyle \vec E_{x} = \frac{\lambda}{4 \pi

\varepsilon_0 a} (\widehat j + \widehat k)$$
$$\displaystyle \vec

E_{net} = \vec E_x + \vec E_y + \vec E_z = \frac{\lambda}{2 \pi

\varepsilon_0 a} (\widehat i + \widehat j + \widehat k)$$
Question 6
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A system consists of a thin charged wire ring of radius r and a very long uniformly charged wire oriented along the axis of the ring, with one of its ends coinciding with the center of the ring. The total charge on the ring is q, and the linear charge density on the straight wire is $$\lambda$$. The interaction force between the ring and the wire is :

A
$$\displaystyle \frac{\lambda q}{4 \pi \varepsilon_0 r}$$

B
$$\displaystyle \frac{\lambda q}{2 \sqrt{2} \pi \varepsilon_0 r}$$

C
$$\displaystyle \frac{2 \sqrt{2} \lambda q}{\pi \varepsilon_0 r}$$

D
$$\displaystyle \frac{4 \lambda q}{\pi \varepsilon_0 r}$$

Solution

Net force $$F_{net} = q E_x$$
Electric field due to a long straight wire of charge density $$\lambda$$ at one of its end is given by, $$E=\dfrac{\lambda}{4 \pi \varepsilon_0 r}$$
$$\displaystyle F = q \dfrac{\lambda}{4 \pi \varepsilon_0} = \dfrac{\lambda q}{4 \pi \varepsilon_0 r}$$
Question 7
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A conic surface is placed in a uniform electric field E as shown in Fig. such that the field is perpendicular to the surface on the side AB. The base of the cone is of radius R, and the height of the cone is h. The angle of the cone is $$\theta $$. Find the magnitude of the flux that enters the cone's curved surface from the left side. Do not count the outgoing flux $$(\theta < 45^{\circ})$$ :

A
$$ER[h \cos \theta + \pi(R/2) \sin \theta ] $$

B
$$ER[h \sin \theta + \pi R/2 \cos \theta ] $$

C
$$ER[h \cos \theta + \pi R \sin \theta ] $$

D
none of these

Solution

Flux entering the cone from side AB will ultimately also pass through areas $$A_1$$ and $$A_2$$. So,
$$\phi =EA_1 cos \theta +EA_2 cos (90-\theta )$$
$$=E\left ( \dfrac{1}{2} 2Rh cos \theta +\dfrac{\pi}{2}R^2 sin \theta \right )$$
$$=ER[h cos \theta + \pi(R/2) sin \theta ]$$
Question 8
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A charge $$Q$$ is divided into two charges $$q$$ and $$Q-q$$. The value of $$q$$ such that the force between them is maximum, is :

A
$$Q$$

B
$$\displaystyle\frac{3Q}{4}$$

C
$$\displaystyle\frac{Q}{2}$$

D
$$\displaystyle\frac{Q}{3}$$

Solution

Let the two charges are separated by a distance $$r$$
$$\therefore$$ Coulomb force $$F = \dfrac{K q (Q-q)}{r^2} = \dfrac{K (qQ-q^2)}{r^2}$$, where $$K = \dfrac{1}{4\pi \epsilon_o}$$

For the force to maximum, $$\dfrac{dF}{dq} = 0$$
$$\therefore$$ $$\dfrac{K}{r^2} \times (Q- 2q) = 0$$
or, $$Q-2q = 0$$ $$\implies q =\dfrac{Q}{2}$$
Question 9
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In a region, the intensity of an electric field is given by $$E = 2i + 3j + k$$ in $$NC^{-1}$$. The electric flux through a surface $$S = 10i \ m^{2}$$ in the region is:

A
$$5\ Nm^{2} C^{-1}$$

B
$$10\ Nm^{2} C^{-1}$$

C
$$15\ Nm^{2} C^{-1}$$

D
$$20\ Nm^{2} C^{-1}$$

Solution

Given: $$E = (2i + 3j + k) NC^{-1}$$ and $$S = 10i \ m^{2}$$
We know that:
$$\phi = E\cdot S$$
$$\phi = (2i + 3j + k) \cdot (10i)$$
$$\phi = 20\ Nm^{2} C^{-1}$$
Question 10
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The spatial distribution of the electric field due to two charges (A, B) is shown in figure. Which one of the following statements is correct ?

A
A is +ve and B is -ve; $$\displaystyle \left| A \right| >\left| B \right| $$

B
A is -ve and B is +ve; $$\displaystyle \left| A \right| =\left| B \right| $$

C
Both are +ve but $$\displaystyle A>B$$

D
Both are -ve but $$\displaystyle A>B$$

Solution

We know that the electric field lines are coming out for positive charge and coming in for negative charge. So A is +ve and B is -ve. As A has more field lines so $$|A|>|B|$$