Two charges 5 × 10−8 C and −3 × 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. - Physics

Advertisements
Advertisements
Numerical

Two charges 5 × 10−8 C and −3 × 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.

Advertisements

Solution

There are two charges,

q1 = 5 × 10−8 C

q2 = −3 × 10−8 C

Distance between the two charges, d = 16 cm = 0.16 m

Consider a point P on the line joining the two charges, as shown in the given figure.

r = Distance of point P from charge q1

Let the electric potential (V) at point P be zero.

Potential at point P is the sum of potentials caused by charges q1 and q2 respectively.

∴ `"V" = q_1/(4piin_0"r") + "q"_2/(4piin_0("d" - "r"))` ........(i)

Where,

`in_0` = Permittivity of free space

For V = 0, equation (i) reduces to

`"q"_1/(4piin_0"r") = -"q"_2/(4piin_0("d" - "r"))`

`"q"_1/"r" = -"q"_2/("d" - "r")`

`(5xx10^-8)/"r" = -(-3 xx 10^-8)/(0.16 - "r")`

`0.16/"r" = 8/5`

∴ r = 0.1 m = 10 cm

Therefore, the potential is zero at a distance of 10 cm from the positive charge between the charges.

Suppose point P is outside the system of two charges at a distance s from the negative charge, where potential is zero, as shown in the following figure.

For this arrangement, potential is given by,

`"V" = "q"_1/(4piin_0"s") + "q"_2/(4 piin_0("s" -"d"))` ......(ii)

For V = 0, equation (ii) reduces to

`"q"_1/(4piin_0"s") = "q"_2/(4piin_0("s" - "d"))`

`"q"_1/"s" = -"q"_2/("s" - "d")`

`(5 xx 10^-8)/"s" = ((-3 xx 10^-8))/(("s" - 0.16))`

`1-0.16/"s" = 3/5`

`0.16/"s" = 2/5`

∴ s = 0.4 m = 40 m

Therefore, the potential is zero at a distance of 40 cm from the positive charge outside the system of charges.

Concept: Potential Due to an Electric Dipole
  Is there an error in this question or solution?
Chapter 2: Electrostatic Potential and Capacitance - Exercise [Page 86]

APPEARS IN

NCERT Physics Class 12
Chapter 2 Electrostatic Potential and Capacitance
Exercise | Q 2.1 | Page 86
NCERT Physics Class 12
Chapter 2 Electrostatic Potential and Capacitance
Exercise | Q 1 | Page 87
Share
Notifications



      Forgot password?
Use app×