Q. 143.6( 13 Votes )

# 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.

Answer :

Potential at any point is given by, V = N m C^{-1}

Electric field generated by a charge, E = N C^{-1}

Where,

q = charge

d = distance from origin

ϵ_{0} = permittivity of space

and, = 9 × 10^{9} N m^{2} C^{-2}

(a) Figure below represents the given situation,

Let the potential at O (midpoint) be V and Electric field be E.

We have,

∴ V = + = 2.4 × 10^{5} N m C^{-1}

and,

E = electric field due to charge 2.5μC – electric field due to charge 1.5μC

∴E =

⇒ E = 4 × 10^{5} N C^{-1}

(b) Figure below represents the given situation,

From the figure point Z is required location and its distance from both the charges is equal and equal to,

d = AZ = BZ =

⇒ d = 0.18 m

and θ =

∴ θ = 56.25^{0}

Thus, Potential and electric field at point Z is given by,

V =

⇒ V = 2 × 10^{5} N m C^{-1}

and,

E =

Where,

E_{1} = electric field generated by 1.5μC charge at Z

E_{2} = electric field generated by 2.5μC charge at Z

and, θ = angle between distance and perpendicular height (as shown above) and,

∴ E = 6.6 × 10^{5} N C^{-1}

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