Question

Q1.Two point charges, QA = +10C and QB = +12C, are separated by a distance r = 5cm. What is the magnitude of the electric force.

Answer 1

Answer:

4.32 × 10^(14) N

Step-by-step explanation:

In between two point charges,

               F = k(q₁q₂)/r²  , k = 9 x 10^9

  (A)<----0.05m---->(B)

  10C                  12C                    

Therefore,

F = (9 × 10^9) × 10 × 12 / (0.05)²

F = (9 × 12 × 10^(10)) / (5 × 10^(-2))²

F = (108 × 10^(10)) / (25 × 10^(-4)

F = 4.32 × 10^(10+4)

F = 4.32 × 10^(14) N

Answer 2

Given :-

Two point charges, QA = +10C and QB = +12C, are separated by a distance r = 5cm

To Find :-

Magnitude

Solution :-

We know that

$\sf F=\dfrac{k(Q_1)(Q_2)}{r^2}$

Where

F = magnitude

k = 9 x 10⁹

Q₁ = First charge

Q₂ = Second charge

r = radius

1 m = 100 cm

5 cm = 5/100 = 0.05 m

$\sf F=\dfrac{9\times 10^9\times10\times12}{(0.05)^2}$

$\sf F=\dfrac{108\times 10^9\times 10^1}{(0.05)^2}$

$\sf F=\dfrac{108 \times 10^{9+1}}{25\times 10^{-4}}$

$\sf F = \dfrac{4.32\times10^{10}}{10^{-4}}$

$\sf F=4.32\times 10^{10}\times10^4$

$\sf F=4.32\times 10^{10+4}$

$\sf F=4.32\times 10^{14}\;N$