Question

Two charges of magnitude 3uc and 4uc are lced at the corners of a and b righted angled triangle abc in whch ab=5 bc=4 and ca=3m. find the magnitude of resultant force expeienced by a charge of 2c at c

Answer 1

$\huge \fcolorbox{red}{pink}{Solution :)}$

Given ,

• Δ ABC is a right angled triangle at C
• 3 uC or 3 × (10)^-6 C charge is located at A
• 4 uC or 4 × (10)^-6 C charge is located at B
• 2 C charge is located at C
• Distance between A and B is 5 m
• Distance between A and C = 3 m
• Distance between B and C = 4 m

We know that , the force between two charged particles is given by

$\large \mathtt{ \fbox{Force = \frac{k( Q_{1}Q_{2})}{ {(r)}^{2} } }}$

Thus ,

$\sf \hookrightarrow Force \: on \: C \: due \: to \: A = \frac{9 \times {10}^{9} \times 3 \times {(10)}^{ - 6} \times 2 }{ {(3)}^{2} } \\ \\ \sf \hookrightarrow Force \: on \: C \: due \: to \: A = 6 \times {(10)}^{3} \: \: newton$

and

$\sf \hookrightarrow Force \: on \: C \: due \: to \: A \: = \frac{9 \times {(10)}^{9} \times 4 \times {(10)}^{ - 6} \times 2 }{ {(4)}^{2} } \\ \\ \sf \hookrightarrow Force \: on \: C \: due \: to \: A \: = \frac{9 \times {(10)}^{3} }{2} \\ \\ \sf \hookrightarrow Force \: on \: C \: due \: to \: A \: = 45 \times {(10)}^{2} \: \: newton$

We know that , the resultant between two perpendicular vectors (A and B) is given by

$\mathtt{\fbox{Resultant = \sqrt{ {(A)}^{2} + {(B)}^{2} } \: }}$

Substitute the known values , we get

$\sf \hookrightarrow Resultant = \sqrt{ {(6 \times {(10)}^{3} )}^{2} + {(45 \times {(10)}^{2}) }^{2} } \\ \\ \sf \hookrightarrow Resultant = \sqrt{36000000 + 20250000} \\ \\ \sf \hookrightarrow Resultant = \sqrt{56250000} \\ \\ \sf \hookrightarrow Resultant = 7500 \: \: newton$

Hence , the required electrostatics force is 7500 newton

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Answer 2

Answer:

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Explanation:

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