Question

Two point charges of 5C and 2C are placed at (2, 3) and (5, 7) respectively. electric force act on 2C, due to 5C is ?

Answer 1

Answer:

Force due to 5C on 2C is 3.6×10⁹ N

Explanation:

Distance between the 2 charges = r = √[ (y₂ - y₁)² + (x₂ - x₁)²]

r = √[ (3 - 7)² + (2 - 5)²]

r = √(4² + 3²)

r = √(16 + 9)

r = √25

r = 5

q₁ = 5C

q₂ = 2C

We know that,

Force = kq₁q₂/r²

F = 9×10⁹×5×2/(5²)

F = 3.6×10⁹ N

Answer 2

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{F_{25}=3.6\times 10^{9}\:N}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt: \implies Two \: point \: charge = 5C \: and \: 2C \\ \\ \tt: \implies Coordinates = (2,3) \: and \: (5,7) \\ \\ \red{\underline \bold{To \: Find:}} \\ \tt: \implies Force \: on \: 2C \: due \: to \: 5C=?$

• According to given question :

$\bold{As \: we \: know \: that} \\ \tt: \implies Distance = \sqrt{ ({ x_{2} - x_{1} )}^{2} + ({ y_{2} - y_{1} )}^{2}} \\ \\ \tt: \implies Distance = \sqrt{(5 - 2) ^{2} + (7 - 3)^{2} } \\ \\ \tt: \implies Distance = \sqrt{ {3}^{2} + {4}^{2} } \\ \\ \tt: \implies Distance = \sqrt{25} \\ \\ \green{\tt: \implies Distance = 5 \: units} \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies F_{25} = \frac{k q_{1} q_{2} }{ {r}^{2} } \\ \\ \tt: \implies F_{25} = \frac{9 \times {10}^{9} \times 2 \times 5}{ {5}^{2} } \\ \\ \tt: \implies F_{25} = \frac{9 \times {10}^{9 + 1} }{25} \\ \\ \tt: \implies F_{25} = \frac{9 \times {10}^{10} }{25} \\ \\ \green{\tt: \implies F_{25} = 3.6 \times {10}^{9} \: N}$