Electric Charges and Fields Test

Result

Electric Charges and Fields Test - 71

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Weekly Quiz Competition

Question 1
1 / -0

The potential at a point $$(x, 0, 0)$$ is given by $$V = \left(\dfrac{1000}{x} + \dfrac{1500}{x^2} + \dfrac{500}{x^3}\right)$$. The intensity of the electric field at $$x = 1$$ will be

A
$$550 V/m$$

B
$$55 V/m$$

C
$$55000 V/m$$

D
$$5500 V/m$$

Solution

$$V=(\cfrac{1000}{x}+\cfrac{1500}{x^{2}}+\cfrac{500}{x^{3}})$$
We know that, $$E=\cfrac{-dv}{dr}$$
So,
$$E=\cfrac{-d(\cfrac{1000}{x}+\cfrac{1500}{x^{2}}+\cfrac{500}{x^{3}})}{dx}$$ at $$x=1$$
$$=-[\cfrac{(-1000)}{x^{2}}-\cfrac{2\times 1500}{x^{3}}-\cfrac{3\times 500}{x^{4}}]_{x=1}$$
$$=1000+3000+1500=5500\,V/m$$
Question 2
1 / -0

Six charges of equal magnitude, $$3$$ positive and $$3$$ negative are to be placed on $$PQRSTU$$ corners of a regular hexagon, such that field at the center is double that of what it would have been if only one $$+$$ve charges is placed at $$R$$

A
$$+,+,+,-,-,-$$

B
$$-,+,+,+,-,-$$

C
$$-,+,+,-,+,-$$

D
$$+,-,+,-,+,-$$

Solution
Question 3
1 / -0

Consider a triangular surface whose vertices are three points having co-ordinate A ( 2a, 0, 0 ), B(0, a, 0), C(0, 0, a). If there is a uniform electric field $$E_0\hat{i} + 2 E_0\hat {j} + 3 E_0 \hat {k} $$ then flux linled to triangular surface ABC is:

A
$$\dfrac{7E_0a^2}{2}$$

B
$$3 E_0a^2$$

C
$$\dfrac{11 E_0a^2}{2}$$

D
Zero

Solution

$$\phi = E_0\times \dfrac{1}{2}a^2 + 2E_0\times \dfrac{1}{2}\times 3E_0\times 3a^2$$
$$= \dfrac{11}{2} E_0a^2$$
Question 4
1 / -0

Fill in the blanks.
A field normal to the plane of a circular wire n turns and radius r which carries a current I is measured on the axis of the coil at small h distance h from the centre of the coil. This is smaller than the field at the centre by a friction ____

A
$$3\dfrac{3h^2}{2r^2}$$

B
$$\dfrac{h^2}{2r^2}$$

C
$$\dfrac{4h^2}{2r^2}$$

D
$$\dfrac{3h^2}{2r^2}$$

Solution

Turns $$=n$$ friction $$=\dfrac { { 3h }^{ 2 } }{ { 2r }^{ 2 } } $$
radius $$=r$$
current $$=I$$
where field at the centre,
$${ B }_{ 1 }=\dfrac { { \mu }_{ 0 } }{ 4\pi } \times \dfrac { 2\pi in }{ r } $$
$$=\dfrac { { \mu }_{ 0 } }{ 2 } \times \dfrac { ni }{ r } $$
field at a distance $$h$$ from the centre
$${ B }_{ 2 }=\dfrac { { \mu }_{ 0 } }{ 4\pi } \dfrac { 2\pi ni{ r }^{ 2 } }{ { \left( { r }^{ 2 }+{ h }^{ 2 } \right) }^{ 3/2 } } $$
$$=\dfrac { { \mu }_{ 0 } }{ 2 } \dfrac { { nir }^{ 2 } }{ { r }^{ 3 }{ \left( 1+\dfrac { { h }^{ 2 } }{ { r }^{ 2 } } \right) }^{ 3/2 } } $$
$$={ B }_{ 1 }{ \left( 1+\dfrac { { h }^{ 2 } }{ { r }^{ 2 } } \right) }^{ 3/2 }$$
$$={ B }_{ 1 }\left( 1-\dfrac { 3 }{ 2 } \dfrac { { h }^{ 2 } }{ { r }^{ 2 } } \right) $$ by Binomial expansion.
Hence $${ B }_{ 2 }<{ B }_{ 1 }$$ by a friction $$=\dfrac { 3 }{ 2 } \dfrac { { h }^{ 2 } }{ { r }^{ 2 } } $$
Question 5
1 / -0

A particle of mass $$m$$ and carrying charge $$-{q}_{1}$$ is moving around a charge $$+{q}_{2}$$ along a circular path of radius $$r$$. Find period of revolution of the charge $$-{q}_{1}$$

A
$$\sqrt { \cfrac { 16{ \pi }^{ 3 }{ \varepsilon }_{ 0 }m{ r }^{ 3 } }{ { q }_{ 1 }{ q }_{ 2 } } } $$

B
$$\sqrt { \cfrac { 8{ \pi }^{ 3 }{ \varepsilon }_{ 0 }m{ r }^{ 3 } }{ { q }_{ 1 }{ q }_{ 2 } } } $$

C
$$\sqrt { \cfrac { { q }_{ 1 }{ q }_{ 2 } }{ 16{ \pi }^{ 3 }{ \varepsilon }_{ 0 }m{ r }^{ 3 } } } $$

D
zero
Question 6
1 / -0

A charge particle $${q}_{1}$$ is at position $$(2, -1, 3)$$. The electrostatic force on another charged particle $${q}_{2}$$ at $$(0, 0, 0)$$ is:

A
$$\dfrac { { q }_{ 1 }{ q }_{ 2 } }{ 56\pi { \in }_{ 0 } } \left( 2\hat { i } -\hat { j } +3\hat { k } \right)$$

B
$$\dfrac { { q }_{ 1 }{ q }_{ 2 } }{ 56\sqrt { 14 } \pi { \in }_{ 0 } } \left( 2\hat { i } -\hat { j } +3\hat { k } \right)$$

C
$$\dfrac { { q }_{ 1 }{ q }_{ 2 } }{ 56\pi { \in }_{ 0 } } \left(\hat { j } -2\hat { i } +3\hat { k } \right)$$

D
$$\dfrac { { q }_{ 1 }{ q }_{ 2 } }{ 56\sqrt { 14 } \pi { \in }_{ 0 } } \left( \hat { j } -2\hat { i } -3\hat { k } \right)$$

Solution
Question 7
1 / -0

Point charges $$+4q$$, $$-q$$ and $$+4q$$ are kept on the $$x-$$axis at points $$x=0,x=a$$ and $$x=2a$$, respectively, then:

A
Only $$q$$ is in stable equilibrium

B
None of the charges are in equilibrium

C
All the charges are in unstable equilibrium

D
All the charges are in stable equilibrium

Solution
Question 8
1 / -0

A particle of charges $${q}_{0}$$ is moved around a charge $$+q$$ along the semicircle path of radius $$r$$ from A to B (see figure). The work done by the Coulomb force is

A
$$\cfrac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \cfrac { q{ q }_{ 0 } }{ r } \quad $$

B
$$\cfrac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \cfrac { 2q{ q }_{ 0 } }{ r } $$

C
$$\cfrac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \cfrac { 2q{ q }_{ 0 } }{ r } .\pi r$$

D
zero

Solution
Question 9
1 / -0

Two positive ions, each carrying a charge q are separated by a distance d. If $$F$$ is the force of repulsion between the ions, the number of electrons missing from each ion will be (c being, the charge of an electron).

A
$$ \dfrac {x \times 10^{-3} \sqrt{F} }{3e} $$

B
$$ \dfrac {x \times 10^{9} \sqrt{F} }{9e^2} $$

C
$$ \dfrac {x \times 10^{-4} \sqrt{F} }{9e^2} $$

D
$$ \dfrac {x \times 10^{5} \sqrt{Fe^2} }{9} $$

Solution

Force of repulsion is given by:

$$F = \dfrac{{k{q^2}}}{{{x^2}}}$$

Let number of electrons missing be$$n$$.

Now $$F = \dfrac{{9 \times {{10}^9} \times {{\left( {ne} \right)}^2}}}{{{x^2}}}$$

$${\left( {ne} \right)^2} = \dfrac{{F{x^2}}}{{9 \times {{10}^9}}}$$

$${n^2} = \dfrac{{F{x^2}}}{{9 \times {{10}^9} \times {e^2}}}$$

$$n = \dfrac{{x\sqrt F }}{{3 \times {{10}^3} \times e}}$$
Question 10
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Two small balls,each having equal positive change $$Q$$ are suspended by two insulating strings of equal length L from a hook fixed to a stand. If the whole set up is transferred to a satellite in orbit around the earth, the tension in equilibrium in each string is equal to:

A
$$0$$

B
$$\dfrac {kQ}{l^2}$$

C
$$\dfrac{kQ^2}{2L^2}$$

D
$$\dfrac{kQ^2}{4L^2}$$

Solution