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3 months ago

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Two small diameter, 10gm dielectric balls can slide freely on a vertical channel each carry a negative charge of 1microcoulomb.

find the separation between the balls if the lower ball is restrained from moving.

1 answer:

kicyunya [3.2K]3 months ago

5 0

Answer:

The separation distance is $d = 0.092 \ m$

Explanation:

The mass of each ball is $m= 10 g = 0.01 \ kg$

The negative charge of each ball is $q_1 =q_2=q = 1 \mu C = 1 *10^{-6} \ C$

Since the lower ball is prevented from moving, it indicates that the net force acting on it equals zero.

This means that the gravitational force exerted on the lower ball balances the electrostatic force, represented as:

$F = \frac{kq_1 * q_2}{d}$

=> $m* g = \frac{kq_1 * q_2}{d}$

where k denotes Coulomb's constant, valued at $k = 9*10^{9} \ kg\cdot m^3\cdot s^{-4}\cdot A^2.$ .

Thus,

$0.01 * 9.8 = \frac{ 9*10^9 *[1*10^{-6} * 1*10^{-6}]}{d}$

$d = 0.092 \ m$

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3 months ago

Read 2 more answers

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#1

The volume of lead measures 100 cm^3

with a density of lead at 11.34 g/cm^3

. Thus, the mass of the lead block equals density multiplied by volume

$m = 100 * 11.34 = 1134 g$

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$W = mg = 1.134* 9.8 = 11.11 N$

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$0.11 = 1000* V * 9.8$

After solving, we find

V = 11.22 cm^3

(ii) This corresponding volume of water exerts the same weight as the buoyant force, resulting in 0.11 N

(iii) The buoyant force measures 0.11 N

(iv) The lead block sinks in water due to its density being greater than that of water.

#2

The buoyant force acting on the lead block counterbalances its weight

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$\rho V g = W$

$13* 10^3 * V * 9.8 = 11.11$

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(iii) The buoyant force is recorded as 11.11 N

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#3

Indeed, an object that has lesser density than a liquid will float; otherwise, it will sink in the liquid.

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