Question

Two identical spheres a and b each having charge 6.5 ✕ 10−7 c are separated by a distance of 50 cm in air. A third uncharged sphere of same size is brought in contact with the first, then brought in contact with the second and afterwards removed from both. What is the new electrical force between the first two spheres?

Answer 1

hope you got the answer.

Answer 2

Answer:

Charge on sphere a, $q_a$ = 6.5 × 10⁻⁷ C

Charge on sphere b, $q_b$ = 6.5 × 10⁻⁷ C

Let the uncharged sphere be c.

Charge on sphere c, $q_c$ = 0 C

When sphere c is brought in contact with the first sphere, i.e., sphere a, then the total charge of the system gets divided equally in spheres a and c.

Total charge on both spheres, q = 6.5 × 10⁻⁷ C

Charge on sphere a and c = (6.5 × 10⁻⁷ )/2 C

                                            = 3.25× 10⁻⁷ C

Now, the sphere c is brought in contact with sphere B. Again the total charge of the system gets divided equally in spheres b and c.

Total charge on both spheres, q' = (3.25× 10⁻⁷ + 6.5 × 10⁻⁷)  C

                                                       = 9.75 × 10⁻⁷ C

Charge on spheres b and c = (9.75 × 10⁻⁷)/2 C

                                              = 4.875 × 10⁻⁷ C

Now, sphere c is removed from both.

New charges on spheres a and b are given by,

New charge on sphere a, $q_a'$ =  3.25 × 10⁻⁷ C

New charge on sphere b, $q_b'$ = 4.875 × 10⁻⁷ C

Electrical force between any two charges is given by,

$F = k \frac{q_1 q_2}{r^2}$

where F = electrical force

k = proportionality constant = 9 × 10⁹ Nm²/C²

q₁ and q₂ are the charges

r = distance between the charges

Here q₁ =  $q_a'$ =  3.25 × 10⁻⁷ C

q₂ =  $q_b'$ = 4.875 × 10⁻⁷ C

r = distance between the charges = 50 cm = 0.5 m

∴ New electrical force between the first two spheres is given by

$F = k \frac{q_a' q_b'}{r^2}$

$F = (9 \times 10^9) \frac{(3.25\times 10^{-7})(4.875\times 10^{-7})}{(0.50)^2}$

$F = (9 \times 10^9) \frac{15.84 \times 10^{-14}}{0.25}$

$F = (9\times 10^9) \ (63.36 \times 10^{-14})\\\\F = 570.24 \times 10^{9-14}\\\\F = 570.24 \times 10^{-5} N\\\\F = 5.7024 \times 10^{-3} N$

Hence, the new electrical force between the first two spheres is 5.7024 × 10⁻³ N.

#SPJ2