Question

4. A charge Q (-5.00 10' C) and two unknown point charges q and q2 are arranged as shown. The electric field at the origin due to these three charges is known to be O. What is the vaide charge q? a. 30° 0.7 m 42 1.4 m
Can someone please help me with this physics problem pertaining to electricity!? I will give a like! Thanks in advance!

Answer 1

Since ,Electric Field is given by,

E = k*q/r^2

where, k = electrostatic constant = 9*10^9

q = charge

r = distance from charge

Direction of Electric field is towards the negative charge and away from positive charge.

Now, due to Q charge, Electric field will be towards North of East, So

E2 = k*Q/r2^2

here, Q = -5*10^-7 C (Negative charge is only used for direction)

r2 = 2.0 m

So,

E2 = (9*10^9)*(5*10^-7)/(2)^2

E2 = 1125 N/C  in 30 deg from +x axis in counter clockwise direction (Because electric field due to negative charge is towards the charge.)

So, E2x = x-component of electric field E2 = E2*cos(30 deg)

E2x = 1125*cos(30 deg)

E2x = 974.29 N/C (In east direction)

Now, due to charge q1, Electric field should be in west direction, for net electric field to be zero at origin. Since due to charge q2 electric field will only be in north/south direction.

for electric field to be in west direction, charge q1 should be positive, as direction of electric field due to positive charge is away from charge. So

E1 = k*q1/r1^2

here, q1 = ??

r1 = 1.4 m

So, E1 = (9*10^9)*q1/(1.4)^2

Given net electric field on origin is equals to zero

So, net horizontal electric field is must be equals to zero

(also charge q2 produce electric field in only y direction)

So,  E1 - E2x = 0

E1 = E2x = 974.29 N/C

now from bold equation,

974.29 = (9*10^9)*q1/(1.4)^2

q1 = 974.29*1.4*1.4/(9*10^9)

q1 = 2.12*10^-7 C

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