Book Store

Download books and chapters from book store.
Currently only available for.
CBSE Gujarat Board Haryana Board

Previous Year Papers

Download the PDF Question Papers Free for off line practice and view the Solutions online.
Currently only available for.
Class 10 Class 12

Electrostatic Potential and Capacitance

A spherical conductor of radius 12 cm has a charge of 1.6 x 10^–7 C distributed uniformly on its surface. What is the electric fields
(a) inside the sphere,
(b) just outside the sphere,
(c) at a point 18 cm from the centre of the sphere?

Given,

$R adius of spherical conductor, r = 0.12 m$

(a) Since, charge resides on the outer surface of the sphere, magnitude of electric field is 0 inside.

(b) Electric field just outside the sphere is given by

$E = \frac{1}{4 {πε}_{0}} \frac{q}{r^{2}} = 9 \times 10^{9} \times \frac{1.6 \times 10^{- 7}}{(0.12)^{2}} = 10^{5} {NC}^{- 1}$

(c) At a point 0.18 m from the centre of the sphere
$E = \frac{1}{4 {πε}_{0}} \frac{q}{r^{2}} = 9 \times 10^{9} \times \frac{1.6 \times 10^{- 7}}{(0.18)^{2}} = 4.4 \times 10^{4} {NC}^{- 1}$

2086 Views

Two charges 2μC and –2μC are placed at points A and B, 6 cm apart.
(a) Identify an equipotential surface of the system.
(b) What is the direction of electric field at every point on this surface?

Given,
$straight q subscript 1 space equals space 2 μC space equals space 2 space cross times space 10 to the power of negative 6 end exponent straight C straight q subscript 2 space equals space minus 2 μc space equals space minus 2 cross times 10 to the power of negative 6 end exponent straight C straight r space equals space 0.06 space straight m$

Given, (a) Potential will be zero due to both charges at eq

(a) Potential will be zero due to both charges at equipotential surface

fraction numerator 1 over denominator 4 πε subscript 0 end fraction open square brackets straight q subscript 1 over straight x plus fraction numerator straight q subscript 2 over denominator left parenthesis 0.06 minus straight x right parenthesis end fraction close square brackets space equals 0

or,

straight q subscript 1 over straight x space equals space minus fraction numerator straight q subscript 2 over denominator left parenthesis 0.06 minus straight x right parenthesis end fraction

or,

fraction numerator 2 cross times 10 to the power of negative 6 end exponent over denominator straight x end fraction space equals space minus fraction numerator open parentheses negative 2 cross times 10 to the power of negative 6 end exponent close parentheses over denominator open square brackets left parenthesis 0.06 right parenthesis minus straight x close square brackets end fraction

or,

straight x space equals space 0.06 space minus space straight x

straight x equals fraction numerator 0.06 over denominator 2 end fraction space equals space 0.03 space straight m

i.e., the plane normal to AB and passing through its mid-point has zero potential everywhere.
(b) The direction of electric field is normal to the plane in the AB direction.

2586 Views

Three capacitors each of capacitance 9 pF are connected in series:
(a) What is the total capacitance of the combination?
(b) What is the potential difference across each capacitor if the combination is connected to a 120 V supply?

Given, three capacitors

straight C subscript 1 space equals space straight C subscript 2 space equals space straight C subscript 3 space equals space 9 straight F space space space space space space equals space 9 space cross times space 10 to the power of negative 12 end exponent straight F

(a) Since the capacitors are connected in series.
Given, three capacitors(a) Since the capacitors are connected in ser

as capacitances are equal so
Given, three capacitors(a) Since the capacitors are connected in ser

1180 Views

A regular hexagon of side 10 cm has a charge 5 μC at each of its vertices. Calculate the potential at the centre of the hexagon.

straight q space equals space 5 space cross times space 10 to the power of negative 6 end exponent straight C straight r space equals space 10 space cm space space equals space 0.1 space straight m

Potential at the centre of the hexagon is given by

straight V space equals space fraction numerator 1 over denominator 4 πε subscript 0 end fraction open square brackets straight q over straight r plus straight q over straight r plus straight q over straight r plus straight q over straight r plus straight q over straight r plus straight q over straight r close square brackets space space space space equals space 6 cross times open parentheses fraction numerator 1 over denominator 4 πε subscript 0 end fraction close parentheses straight q divided by straight r

or,

straight V space equals space fraction numerator 6 cross times 9 cross times 10 to the power of 9 cross times 5 cross times 10 to the power of negative 6 end exponent over denominator 0.1 end fraction space equals space 2.7 space cross times space 10 to the power of 6 space straight V.

2320 Views

Two charges 5 x 10^–8C and –3 x 10^–8 C are located 16 cm apart. At what point(s) on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.

The other possibility of point where total potential due to q₁ and q₂ are zero may be that point lies out side the segment joining q₁ and q₂Let required point P lies x from