Science|Math|Technology
Q: A positive charge is moved from a low potential point A to a high potential point B. Then the electric potential energy:
(a) Increases
(b) decreases
(c) will remain the same
(d) nothing definite can be predicted
Ans: (a)
Sol: $\large \Delta U = q (V_B – V_A ) $
$\large \Delta U = q \Delta V $
As ΔV increases , so ΔU increases .
Q: Three charges 2q , -q ,-q are located at the vertices of an equilateral triangle. At the centre of the triangle:
(a) The field is zero but potential is non – zero
(b) The field is non – zero but potential is zero
(c) Both field and potential are zero
(d) Both field and potential are non – zero
Ans: (b)
Sol: Let ABC is an equilateral triangle having centre O . Where OA = OB = OC = x (say)
Let three charges 2q , -q ,-q are located at the vertices A , B , C respectively .
Electric field at O due to 2q is $E_{OA} = \frac{1}{4 \pi \epsilon_0} \frac{2q}{x^2}$ (along AO produced)
Electric field at O due to B(-q ) is $E_{OB} = \frac{1}{4 \pi \epsilon_0} \frac{q}{x^2}$ (along OB)
and , Electric field at O due to C(-q) is $E_{OB} = \frac{1}{4 \pi \epsilon_0} \frac{q}{x^2}$ (along OC)
Now , find the resultant field these three field we can see that Electric field at O is not zero .
But , Electric Potential at O is ,
$\large V = \frac{1}{4 \pi \epsilon_0}\frac{2q}{x} + \frac{1}{4 \pi \epsilon_0}\frac{-q}{x} + \frac{1}{4 \pi \epsilon_0}\frac{-q}{x}$
V = 0
Q: Two equal -ve charges -q are fixed at the point (0,a) and (0,-a) on the y – axis. A positive charge Q is released from rest at the point (2a ,0) on the x- axis. The charge will
(a) Execute SHM about the origin
(b) Move to the origin and remain at rest
(c) Move to infinity
(d) Execute oscillatory but not SHM
Ans: (d)
Q: If two like charges of magnitude 1 × 10-9 coulomb and 9 × 10-9 coulomb are separated by a distance of 1 metre, then the point on the line joining the charges, where the force experienced by a charge placed at that point is zero, is:
(a) 0.25 m from the charge 1 × 10-9 coulomb
(b) 0.75 m from the charge 9 × 10-9 coulomb
(c)Both 1 and 2
(d) At all points on the line joining the charges
Ans: (c)
Sol: Let AB be the line joining the charges A (1 × 10-9 C) and B (9 × 10-9 C) . Let a charge q be placed at Point C & force acting on it is zero and AC = x , therefore BC = (1-x)
$F_{CA} = F_{CB}$
$\large \frac{1}{4 \pi \epsilon_0} \frac{q \times 1 \times 10^{-9}}{x^2}= \frac{1}{4 \pi \epsilon_0} \frac{q \times 9 \times 10^{-9}}{(1-x)^2} $
$\large \frac{1}{x^2} = \frac{9}{(1-x)^2} $
$\large \frac{1}{x} = \frac{3}{1-x} $
3x = 1-x ⇒ 4x = 1
x = 0.25 m
Q: A proton has a mass of 1.67 × 10-27 kg and charge 1.6 × 10-19 coulomb. If the proton is to be accelerated through a potential difference of one million volt, then the KE is
(a) 1.6 × 10-15 J
(b) 1.6 × 10-13 J
(c) 1.6 × 10-25 J
(d) 3.2 × 10-13 J
Ans: (b)
Sol: $\large KE = \frac{1}{2} m v^2 = e V$
$\large KE = 1.6 \times 10^{-19} \times 10^6$
= 1.6 × 10-13 J
Q: Two concentric thin metallic spheres of radii R1 and R2 (R1 >R2 ) bear charges Q1 and Q2 respectively. Then the potential at r between R1 and R2 will be 1/(4πε0 ) times
(a) $\large \frac{Q_1 + Q_2}{r} $
(b) $\large \frac{Q_1}{R_1}+ \frac{Q_2}{r} $
(c) $\large \frac{Q_1}{R_1}+ \frac{Q_2}{R_2} $
(d) $\large \frac{Q_2}{R_1}+ \frac{Q_1}{R_2} $
Ans: (b)
Sol: Let P be the point & O be the centre of concentric spheres , OP = r (say)
Potential at P = Potential at P due to Q1 + Potential at P due to Q2
$\large V = \frac{1}{4\pi \epsilon_0} \frac{Q_1}{R_1} + \frac{1}{4\pi \epsilon_0} \frac{Q_2}{r} $
$\large V = \frac{1}{4\pi \epsilon_0} ( \frac{Q_1}{R_1} + \frac{Q_2}{r}) $
Q: The force between two charges situated in air is F. The force between the same charges if the distance between them is reduced to half and they are situated in a medium having dielectric constant 4 is
(a) F/4
(b) 4F
(c) 16F
(d) F
Ans: (d)
Sol: $\large F = \frac{1}{4\pi \epsilon_0} \frac{q_1 q_2}{r^2}$ …..(i)
In medium ,
$\large F’ = \frac{1}{4\pi \epsilon_0 \epsilon_r} \frac{q_1 q_2}{(r/2)^2} $ …..(ii)
On dividing ,
$\large \frac{F’}{F} = \frac{4}{\epsilon_r} = \frac{4}{4}= 1$
F’ = F