four equal charges of magnitude 2 micro coulomb each are placed on the corners of a square abcd having side length 2 cm . find the net force acting on the charge placed at A help meee
Answers
Answer:
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First of all, we have to draw the figure(see attached document).
Then we can apply Coulomb’s law to calculate electric force on 1 µC due to other 4 point chargers.
F = (1/4πε0)*(q1q2/r²)
Where, q1 = first charge
q2 = Second charge
r = distance between both charge
ε0= permittivity of free space
But, for this question ε0 is a constant. Therefore 1/4πε0 is also a constant.
Then we can change above equation as below,
F = K*(q1q2/r^{2} )F=K∗(q1q2/r
2
)
F_{A} = K*(2/OA^{2} )F
A
=K∗(2/OA
2
) -------- (1)
F_{B} = K*(-5/OB^{2} )F
B
=K∗(−5/OB
2
) --------(2)
F_{C} = K*(2/OC^{2} )F
C
=K∗(2/OC
2
) ---------(3)
F_{D} = K*(-5/OD^{2} )F
D
=K∗(−5/OD
2
) --------(4)
We know that OA= OB=OC=OD
Hence \begin{gathered}F_{A} = F_{C} \\F_{B} = F_{D}\end{gathered}
F
A
=F
C
F
B
=F
D
But Fa, Fc & Fb,Fd are in opposite directions. (see document)
Hence sum of all forces will =0
Answer :force on a charge of 1uc = 0
this is your answer I hope it helps you
