Question

Find out force on 2C charge at (3,4)due to -5C charge at (6,8)

Answer 1

$\huge \underline {\underline{ \mathfrak{ \green{Ans}wer \colon}}}$

Given :

• 1st Charge (Q1) = 2 C
• 2nd Charge (Q2) = -5 C
• Q1 is placed at (3,4)
• Q2 is placed at (6,8)

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To Find :

• Electrostatic Force between the charges

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Solution :

The points are (3,4) and (6,8)

Where,

• x1 = 3
• y1 = 4
• x2 = 6
• y2 = 8

Use distance Formula :

$\large{\boxed{\sf{d \: = \: \sqrt{(x_2 \: - \: x_1)^2 \: + \: (y_2 \: - \: y_1)^2}}}} \\ \\ \implies {\sf{d \: = \:\sqrt{ (6 \: - \: 3)^2 \: + \: (8 \: - \: 4)^2}}} \\ \\ \implies {\sf{d \: = \: \sqrt{ (3)^2 \: + \: (4)^2}}} \\ \\ \implies {\sf{d \: = \: \sqrt{ 9 \: + \: 16}}} \\ \\ \implies {\sf{d \: = \: \sqrt{25}}} \\ \\ \implies {\sf{d \: = \: 5 \: units}} \\ \\ \underline{\sf{\therefore \: Distance \: between \: the \: points \: is \: 5 \: units}}$

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We know that Coulomb's Law :

$\large{\boxed{\rm{F \: = \: k \dfrac{Q_1 Q_2}{d^2}}}} \\ \\ \implies {\sf{F \: = \: 9 \: \times \: 10^9 \: \dfrac{2 \: \times \: -5}{(5)^2}}} \\ \\ \implies {\sf{F \: = \: \dfrac{9 \: \times \: 10 ^9 \: \times \: -10}{25}}} \\ \\ \implies {\sf{F \: = \: \dfrac{-90 \: \times \: 10^9}{25}}} \\ \\ \implies {\sf{F \: = \: \dfrac{-90 \: \times \: 100 \: \times \: 10^7}{25}}} \\ \\ \implies {\sf{F \: = \: -90 \: \times \: 4 \: \times \: 10^7}} \\ \\ \implies {\sf{F \: = \: -360 \: \times \: 10^6}} \\ \\ \implies {\sf{F \: = \: -36 \: \times \: 10^7}} \\ \\ {\underline{\sf{\therefore \: Force \: between \: the \: charges \: is \: -36 \: \times \: 10^7 \: N}}}$

Answer 2

$\large \bigstar\boxed{\bf \red{ GIVEN}} \bigstar \\ \\</p><p></p><p>\bf {Q}_{1} = 2 C \\</p><p>\bf{Q}_{2} = -5 C \\</p><p>\bf{Q}_{1} \: is \: placed \: at \: (3,4) \\</p><p>\bf{Q}_{2} \: is \:placed \:at \: (6,8)$

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$\large \bigstar\boxed{\bf \red{TO \: FIND}} \bigstar \\ \\ </p><p>\bf Electrostatic \: Force \:between\: the \:charges$

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$\large \bigstar\boxed{\bf \red{SOLUTION}} \bigstar$

The points are (3,4) and (6,8)

Where,

$\bf{x}_{1} = 3 \\</p><p>\bf{y}_{1} = 4 \\</p><p>\bf{x}_{2} = 6 \\ </p><p>\bf{y}_{2} = 8$

Use distance Formula :

$\large \boxed{\sf \green{d \: = \: \sqrt{(x_2 \: - \: x_1)^2 \: + \: (y_2 \: - \: y_1)^2}}} \\ \\ \implies {\sf{d \: = \:\sqrt{ (6 \: - \: 3)^2 \: + \: (8 \: - \: 4)^2}}} \\ \\ \implies {\sf{d \: = \: \sqrt{ (3)^2 \: + \: (4)^2}}} \\ \\ \implies {\sf{d \: = \: \sqrt{ 9 \: + \: 16}}} \\ \\ \implies {\sf{d \: = \: \sqrt{25}}} \\ \\ \boxed{\sf{d \: = \: 5 \: units}} \\ \\ \underline{\sf{\therefore \: Distance \: between \: the \: points \: is \: 5 \: units}}$

$\large \boxed{ \bf \blue{F \: = \: k \dfrac{Q_1 Q_2}{d^2}}} \\ \\ \implies {\sf{F \: = \: 9 \: \times \: 10^9 \: \dfrac{2 \: \times \: -5}{(5)^2}}} \\ \\ \implies {\sf{F \: = \: \dfrac{9 \: \times \: 10 ^9 \: \times \: -10}{25}}} \\ \\ \implies {\sf{F \: = \: \dfrac{9 \: \times \: 10^8}{25}}} \\ \\ \implies {\sf{F \: = \: \dfrac{9 \: \times \: 100 \: \times \: 10^6}{25}}} \\ \\ \implies {\sf{F \: = \: 9 \: \times \: 4 \: \times \: 10^6}} \\ \\ \boxed {\sf \orange{F \: = \: 36 \: \times \: 10^6}} \\ \\ {\underline{\sf{\therefore \: Force \: between \: the \: charges \: is \: 36 \: \times \: 10^6 \: N}}}$