Question

Two charges of 2micro c but opposite in sign are placed 10 CM apart calculate electric field at point 10 cm apart from mid point on axial line

Answer 1

From my calculations the answer is 6.4×10^6 N/C

Explanation:

360/5.625×10^-5=6.4×10^6

Appreciate what you've done,by the way

Answer 2

Answer:

The electric field from the mid point on the axial line is 0.64×10⁷N/C.

Explanation:

The opposite charges on the axial line will behave like a dipole. So, the electric field of the dipole on the axial line is given as,

$E=\frac{kq_1}{r_1} +\frac{kq_2}{r_2}$    (1)

Where,

E=electric field on the axial line of the dipole

q₁,q₂=charges on the dipole

k=coulombs constant=9×10⁹kg⋅m³⋅s⁻²⋅C⁻²

r₁,r₂=distance of the point from the respective charges

From the question we have,

q₁=2μC=2×10⁻⁶C

q₂=-2μC=-2×10⁻⁶C

r₁=5cm=5×10⁻²m

r₂=15cm=15×10⁻²m

By inserting the entities in equation (1) we get;

$E=k(\frac{2\times 10^-^6}{5\times 10^-^2} +\frac{-2\times 10^-^6}{15\times 10^-^2})$

$E=\frac{9\times 10^9\times 2\times 10^-^6}{(5\times 10^-^2)^2} (1-\frac{1}{3^2})$

$E=\frac{18\times 10^3}{25\times 10^-^4} \times \frac{8}{9}$

$E=0.64\times 10^7\ N/C$

Hence, the electric field from the mid point on the axial line is 0.64×10⁷N/C.

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