11. Suppose the charge of a proton and an electron differ slightly. One of them is – e, the other is (e + ∆e). If the net of electrostatic force and gravitational force between two hydrogen atoms placed at a distance d (much greater than atomic size) apart is zero, then ∆e is of the order of [Given mass of hydrogen mh = 1.67 × 10^–27 kg]
    

1. 1. ​10^–23 C

   
   
   

   2. 10^–37 C

   
   
   

   3. 10^–47 C

   
   
   

   4. 10^–20 C

   
   
   

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1. View Hint View Answer Discuss in Forum

   According to question, the net electrostatic force (F_E) = gravitational force  (F_G)
   ​F_E = F_G
   

   or	1		∆e2	=	Gm2
   	4πε0		d2		d2

   
   

   ⇒ ∆e = m
   		G
   		K

   
   

   ∵		1	= k = 9 × 109
   		4πε0

   
   

   = 1.67 × 10–27
   		6.67 × 10–11
   		9 × 109

   
   ∆e ≈ 1.436 × 10^–37 C

   Correct Option: B

   According to question, the net electrostatic force (F_E) = gravitational force  (F_G)
   ​F_E = F_G
   

   or	1		∆e2	=	Gm2
   	4πε0		d2		d2

   
   

   ⇒ ∆e = m
   		G
   		K

   
   

   ∵		1	= k = 9 × 109
   		4πε0

   
   

   = 1.67 × 10–27
   		6.67 × 10–11
   		9 × 109

   
   ∆e ≈ 1.436 × 10^–37 C

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12. Two identical charged spheres suspended from a common point by two massless strings of lengths l, are initially at a distance d (d << l) apart because of their mutual repulsion. The charges begin to leak from both the spheres at a constant rate. As a result, the spheres approach each other with a velocity v. Then v varies as a function of the distance x between the spheres, as :
    
    
    

1. 1. v ∝ x^{1 / 2}

   
   
   

   2. v ∝ x

   
   
   

   3. v ∝ x^{-1 / 2}

   
   
   

   4. v ∝ x^-1

   
   
   

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1. View Hint View Answer Discuss in Forum

   From figure tan θ =	Fc	; θ
   	mg

   
   

   	kq2	=	x
   	x2 mg		2ℓ

   
   
   or​ x³ ∝ q²  …(1) ​
   or​ x^{3/ 2} ∝ q …(2) ​
   Differentiating eq. (1) w.r.t. time
   

   3x2	dx	∝ 2q	dq	but	dq	is constant ​
   	dt		dt		dt

   
   so x²(v) ∝ q Replace q from eq. (2) ​
   x²(v) ∝ x^3/2 or v ∝ x^–1/2

   Correct Option: C

   From figure tan θ =	Fc	; θ
   	mg

   
   

   	kq2	=	x
   	x2 mg		2ℓ

   
   
   or​ x³ ∝ q²  …(1) ​
   or​ x^{3/ 2} ∝ q …(2) ​
   Differentiating eq. (1) w.r.t. time
   

   3x2	dx	∝ 2q	dq	but	dq	is constant ​
   	dt		dt		dt

   
   so x²(v) ∝ q Replace q from eq. (2) ​
   x²(v) ∝ x^3/2 or v ∝ x^–1/2

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13. Two pith balls carrying equal charges are suspended from a common point by strings of equal length. The equilibrium separation between them is r. Now the strings are rigidly clamped at half the height. The equilibrium separation between the balls now become​​
    

1. 1. 		r
      		3 √2

   
   
   

   2. 		2r
      		√3

   
   
   

   3. 		2r
      		3

   
   
   

   4. 		r			2
      		√2

   
   
   

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1. View Hint View Answer Discuss in Forum

   
   

   From figure,  tan θ =	Fe	⇒	r / 2	=	kq2
   	mg		y		r2 / mg

   
   

   [ ∵ F =	kq2	from coulomb’s law]
   	r2

   
   

   ⇒ r3 ∝ y ⇒ r'3 ∝	y	⇒	r'	=	1
   	2		r		21 / 3

   
   

   ⇒ r' =	r
   	3√2

   Correct Option: A

   
   

   From figure,  tan θ =	Fe	⇒	r / 2	=	kq2
   	mg		y		r2 / mg

   
   

   [ ∵ F =	kq2	from coulomb’s law]
   	r2

   
   

   ⇒ r3 ∝ y ⇒ r'3 ∝	y	⇒	r'	=	1
   	2		r		21 / 3

   
   

   ⇒ r' =	r
   	3√2

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14. A charge ‘q’ is placed at the centre of the line joining two equal charges ‘Q’. The system of the three charges will be in equilibrium if ‘q’ is equal to​​

1. 1. Q/2

   
   
   

   2. – Q/4

   
   
   

   3. Q/4 ​

   
   
   

   4. – Q/2

   
   
   

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1. View Hint View Answer Discuss in Forum

   The system of three charges will be in equilibrium.
   
   For this, force between charge at A and B + force between charge at point O and either at  A or B is zero.
   

   i.e. ,	KQ2	+	KQq	= 0
   	r2		(r / 2)2

   
   By solving we get,
   

   q = -	Q
   	4

   Correct Option: B

   The system of three charges will be in equilibrium.
   
   For this, force between charge at A and B + force between charge at point O and either at  A or B is zero.
   

   i.e. ,	KQ2	+	KQq	= 0
   	r2		(r / 2)2

   
   By solving we get,
   

   q = -	Q
   	4

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15. An electron is moving round the nucleus of a hydrogen atom in a circular orbit of radius r. The Coulomb force F^→ between the two is​

1. 1. K	e2	r→
      	r3

   
   
   

   2. K	e2	r̂
      	r2

   
   
   

   3. -K	e2	r̂
      	r3

   
   
   

   4. -K	e2	r→
      	r3

      
      

      	where K =	1
      		4πε0

   
   
   

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1. View Hint View Answer Discuss in Forum

   Charges (–e) on electron and (e) on proton exert a force of attraction given by ​
   

   Force = (K)	(-e)(e)	r̂ =	-Ke2	ṙ		∵ r̂ =	r→
   	r2		r3				| r |

   
   Note : Magnitude of Coulomb force is given by
   

   F =	1	.	q1q2
   	4πε0		r2

   
   

   but in vector form F→ =	1	.	q1q2	r→
   	4πε0		r2

   
   

   Correct Option: D

   Charges (–e) on electron and (e) on proton exert a force of attraction given by ​
   

   Force = (K)	(-e)(e)	r̂ =	-Ke2	ṙ		∵ r̂ =	r→
   	r2		r3				| r |

   
   Note : Magnitude of Coulomb force is given by
   

   F =	1	.	q1q2
   	4πε0		r2

   
   

   but in vector form F→ =	1	.	q1q2	r→
   	4πε0		r2

   
   

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