Question

(a). What is the direction of torque acting on electric dipole held at an angle with uniform external electric field?

(b). An electric dipole of length 10cm having charges ±6× 10⁻³C, placed at 30˚ w.r.t. a uniform electric field experiences a torque of magnitude 6√3 Nm. 

Calculate : 

i. Magnitude of electric field, 

ii. Potential energy of dipole, 

(c). Calculate electrical capacity of earth. 

(d). Depict the equipotential surfaces for 

i. A single isolated change 

ii. A system of two identical point charges placed distance ‘d’ apart.

Answer 1

(a). Torque, 

τ = pe sinθ, 

Which can be written in vector form as : 

\(\vec τ= \vec p \times \vec E\)

Therefore, 

The direction of torque \(\vec τ\) is perpendicular to both \(\vec p\)and \(\vec E\). It is determined by right handed screw rule.

(b) Here,

2a = 10cm = 10⁻¹m 

q = ±6 × 10⁻³C 

θ = 30° 

\(\vec τ\) = 6√3 Nm 

E = ? 

U = ?

(i) From τ = PE sinθ 

= q (2a)E sinθ 

(ii) U = - pE cosθ

= - q(2a) E cosθ

= - 6× 10⁻³ × 10⁻¹ × 2√3 × 10⁴ × cos30°

= - 6 × 10⁻⁴ × 2√3 × 10⁴ × \(\frac{\sqrt 3}{2}\) 

= -18J.

(c) Here, 

r = 6400 km 

= 6.4× 10⁶m

As, 

C = 4π∈₀r

= \(\frac{1}{9\times10^{9}}\) × (6.4 × 10⁶) 

= 0.711× 10⁻³ farad 

= 0.711 × 10⁻³ × 10⁶µf 

= 711 µf

Thus, 

The capacity of earth, which is the biggest sphere we come across is only 711µf. 

Therefore, in routine, 

Capacitance will be in nano-farad (nf) and picofarad (pf). Farad is too big a unit of capacitance.

(d). (i)

(ii).