Question

Two insulated copper spheres A and B have their centres separated by distance of 50cm. what is the mutual force of electrostatic
repulsion if the charge on each is 6.5x10^-7. The radii of A and B are negligible compared to the distance of separation. Also compare attraction if each mass is 0.5kg​

Answer 1

Answer:

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Explanation:

(a) Charge on sphere A, qA = 6.5 × 10−7 C Charge on sphere B, qB = 6.5 × 10−7 C Distance between the spheres, r = 50 cm = 0.5 m

Force of repulsion between the two spheres Therefore, the force between the two spheres is 1.52 × 10−2 N.

(b) After doubling the charge, Charge on sphere A, qA = 1.3 × 10−6 C Charge on sphere B, qB =

1.3 × 10−6 C

• The distance between the spheres is halved. ∴=0.5/2=0.25
• $\frac{0.5}{2}$

Force of repulsion between the two spheres, Therefore, the force between the two spheres is 0.243 .

Answer 2

Solution:

As per the given data,

• Distance between the two spheres= 50 cm = 0.5 m
• Q₁ = Q₂ = 6.5 x 10⁻⁷ C
• The radii of both the charges are assumed to be negligible.

The electrostatic force of attraction acting between two charges kept at a distance r from each other is given by,

$\implies\sf{F = \dfrac{KQ_1Q_2}{r^2}}$

Now let's substitute the given values,

$\implies\sf{F = \dfrac{9 \times 10^9\times 6.5 \times 10^{-7}\times 6.5 \times 10^{-7} }{(0.5)^2}}$

$\implies\sf{F = \dfrac{380.25 \times 10^{-5} }{0.25}}$

$\implies\boxed{\sf{F = 1521 \times 10^{-5} N }}$

Now,

• Mass of each copper sphere  = 0.5 kg

We know that,

The electrostatic force of attraction doesn't depend upon the mass of the sphere.

Hence,

$\implies\boxed{\sf{F = F' = 1521 \times 10^{-5} N }}$

The ratio of both the forces,

$\implies\boxed{\sf{\dfrac{F}{F'} = 1 }}$