16. Two charges q₁ and q₂ are placed 30 cm apart, as shown in the figure. A third charge q3 is moved along the arc of a circle of radius 40 cm from C to D. The change in the potential energy of the system is (q₃/4πε₀)k where k is
    

1. 1. 8q₁

   
   
   

   2. 6q₁

   
   
   

   3. 8q₁

   
   
   

   4. 6q₁

   
   
   

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1. View Hint View Answer Discuss in Forum

   We know that potential energy of discrete system of charges is given by ​ ​
   

   U =	1		q1q2	+	q2q3	+	q3q1
   	4πε0		q12		q23		q31

   
   According to question, ​
   

   Uinitial =	1		q1q2	+	q2q3	+	q3q1
   	4πε0		0.3		0.5		0.4

   
   

   Ufinal =	1		q1q2	+	q2q3	+	q3q1
   	4πε0		0.3		0.1		0.4

   
   

   Ufinal - Uinitial =	1		q1q2	-	q2q3
   	4πε0		0.1		0.4

   
   

   =	1	[10q2q3 - 2q2q3] =	q3	(8q2)
   	4πε0		4πε0

   Correct Option: C

   We know that potential energy of discrete system of charges is given by ​ ​
   

   U =	1		q1q2	+	q2q3	+	q3q1
   	4πε0		q12		q23		q31

   
   According to question, ​
   

   Uinitial =	1		q1q2	+	q2q3	+	q3q1
   	4πε0		0.3		0.5		0.4

   
   

   Ufinal =	1		q1q2	+	q2q3	+	q3q1
   	4πε0		0.3		0.1		0.4

   
   

   Ufinal - Uinitial =	1		q1q2	-	q2q3
   	4πε0		0.1		0.4

   
   

   =	1	[10q2q3 - 2q2q3] =	q3	(8q2)
   	4πε0		4πε0

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17. An electric dipole has the magnitude of its charge as q and its dipole moment is p. It is placed in uniform electric field E. If its dipole moment is along the direction of the field, the force on it and its potential energy are respectively

1. 1. q.E and max.

   
   
   

   2. 2 q.E and min.

   
   
   

   3. q.E and p.E

   
   
   

   4. zero and min.

   
   
   

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1. View Hint View Answer Discuss in Forum

   When the dipole is in the direction of field then net force is qE + (– qE) = 0
   
   ​and its potential energy is minimum ​= – P.E. = – qaE

   Correct Option: D

   When the dipole is in the direction of field then net force is qE + (– qE) = 0
   
   ​and its potential energy is minimum ​= – P.E. = – qaE

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18. A bullet of mass 2 g is having a charge of 2µC. Through what potential difference must it be accelerated, starting from rest, to acquire a speed of 10 m/s?

1. 1. 50 V

   
   
   

   2. 5 kV

   
   
   

   3. 50 kV

   
   
   

   4. 5 V

   
   
   

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1. View Hint View Answer Discuss in Forum

   qV =	1	mv²
   	2

   
   

   ⇒ 2 × 10-6 × V =	1	×	2	× 10 × 10
   	2		1000

   
   ∴  V = 50 kV

   Correct Option: C

   qV =	1	mv²
   	2

   
   

   ⇒ 2 × 10-6 × V =	1	×	2	× 10 × 10
   	2		1000

   
   ∴  V = 50 kV

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19. Each corner of a cube of side l has a negative charge, –q. The electrostatic potential energy of a charge q at the centre of the cube is

1. 1. 4q²

      √2πε0l

   
   
   

   2. √2q²

      4πε0l

   
   
   

   3. 4q²

      √2πε0l

   
   
   

   4. -	4q²
      	√3πε0l

   
   
   

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1. View Hint View Answer Discuss in Forum

   ​Length of body diagonal = ​√3l
   ∴ Distance of centre of cube from each corner ​
   

   r =	√3	l
   	2

   
   ​P.E. at centre ​
   = 8 × Potential Energy due to A ​
   

   = 8 ×	√3	l
   	2

   
   

   = 8 ×	1	× 2 × q × (-q) =	-4q²
   	4πε0√3l		√3πε0l

   
   

   Correct Option: D

   ​Length of body diagonal = ​√3l
   ∴ Distance of centre of cube from each corner ​
   

   r =	√3	l
   	2

   
   ​P.E. at centre ​
   = 8 × Potential Energy due to A ​
   

   = 8 ×	√3	l
   	2

   
   

   = 8 ×	1	× 2 × q × (-q) =	-4q²
   	4πε0√3l		√3πε0l

   
   

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20. A particle of mass m and charge q is placed at rest in a uniform electric field E and then released. The kinetic energy attained by the particle after moving a distance y is

1. 1. qEy²

   
   
   

   2. qE²y

   
   
   

   3. qEy

   
   
   

   4. q²Ey

   
   
   

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1. View Hint View Answer Discuss in Forum

   K.E. = Force × distance  = qE.y

   Correct Option: C

   K.E. = Force × distance  = qE.y

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