Two tiny spheres carrying charges 1.5 μC and 2.5 μC are located 30 cm apart. Find the potential and electric field: (a) at the mid-point of the line joining the two charges, and (b) at a point 10 cm from this midpoint in a plane normal to the line and passing through the mid-point.
Answers
(a) two tiny spheres carrying charges q1 = 1.5 uC and q2 = 2.5uC are located r = 30cm apart.
so, potential at midpoint of the line joining the two charges, V = kq1/(r/2) + kq2/(r/2)
= 9 × 10^9 × 1.5 × 10^-6/(30 × 10^-2/2) + 9 × 10^9 × 2.5 × 10^-6/(30 × 10^-2/2)
= 9 × 10^9 × 10^-6/10^-2 [ 1.5/15 + 2.5/15 ]
= 9 × 10^5 × (4/15)
= 2.4 × 10^5 Volts
hence, potential at midpoint , V = 2.4 × 10^5 volts.
similarly, electric field at midpoint , E = kq1/(r/2)² + kq2/(r/2)²
= 9 × 10^9 × (1.5 × 10^-6)/(30/2 × 10^-2)² + 9 × 10^9 × (2.5 × 10^-6)/(30/2 × 10^-2)²
= 4 × 10^5 V/m
(b) let point P is located 10cm perpendicularly above the midpoint of the line joining the two charges.
seperation between charge and point P,. r = √(15² + 10²) ≈ 18cm or 18 × 10^-2 m
so, potential at point P, Vp = (kq1/r)+(kq2/r)
= 9 × 10^9 × 1.5 × 10^-6/(18 × 10^-2) + 9 × 10^9 × 1.5 × 10^-6/(18 × 10^-2)
= 2 × 10^5 volts
horizontal component of electric field at point P = (kq1/r²)cosα - (kq2/r²)cosα
where cosα = b/h = 15/18
= (9 × 10^9 × 1.5 × 10^-6)/(18 × 10^-2)² × 15/18 - (9 × 10^9 × 2.5 × 10^-6)/(18 × 10^-2)² × 15/18
= -2.3 × 10^5 V/m
vertical component of electric field at point P = (kq1/r²)sinα + (kq2/r²)sinα
where, sinα = 10/18
= (9 × 10^9 × 1.5 × 10^-6)/(18 × 10^-2)² × 10/18 + (9 × 10^9 × 2.5 × 10^-6)/(18 × 10^-2)² × 10/18
= 6.2 × 10^5 V/m
hence, electric field at point P = (-2.3 × 10^5)i + (6.2 × 10^5)j
so, magnitude of electric field at point P = √{(-2.3 × 10^-5)² + (6.2 × 10^-5)²}
= 6.6 × 10^-5 V/m directed at an angle tanα = Ex/Ey = (6.2 × 10^5)/(-2.3 × 10^5) = - 2.6956 ⇒α = -69.6° with horizontal in -ve x - direction.