two charges of magnitude 3uc and 4uc Columbus kept at (8, 4, 2) and)(0, - 2, 2) respectively find the force between them in vector form
Answers
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 ,
\begin{gathered} \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 \end{gathered}
↪ForceonCduetoA=
(3)
2
9×10
9
×3×(10)
−6
×2
↪ForceonCduetoA=6×(10)
3
newton
and
\begin{gathered} \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\end{gathered}
↪ForceonCduetoA=
(4)
2
9×(10)
9
×4×(10)
−6
×2
↪ForceonCduetoA=
2
9×(10)
3
↪ForceonCduetoA=45×(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
\begin{gathered} \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\end{gathered}
↪Resultant=
(6×(10)
3
)
2
+(45×(10)
2
)
2
↪Resultant=
36000000+20250000
↪Resultant=
56250000
↪Resultant=7500newton
Hence , the required electrostatics force is 7500 newton