Problem 4: Two charges of +2 x 10-7C and -5 x 10-7C are placed at a distance of 50cm from each other. Find a point on the line joining the charges at which the electric field is zero.
Electric Field:
It is a region around a charge body in which another charge experiences an electric force.
Intensity of Electric Field:
The force experienced by the test charge at a point per unit charge is known as intensity of Electric Field. Mathematically:
Solution:
Step: 1 Overview
Electric Field and Its Intensity:Electric Field:
It is a region around a charge body in which another charge experiences an electric force.
Intensity of Electric Field:
The force experienced by the test charge at a point per unit charge is known as intensity of Electric Field. Mathematically:
E=F/qo
We know that: F=Kqqo/r2
E=(Kqqo/r2)/qo
E=Kq/r2
Step: 2 Calculation
Intensity of field at P will be zero if and only if the intensity due to both charges is same at that point because of different charges.
Intensity at point P due to charge q1= +2 x 10-7C:E1=Kq1/X2Intensity at point P due to charge q2= -5 x 10-7C:
E2=Kq2/(X+0.5)2Since both are equal at point P,
E1 = E2dividing K both sides,
Kq1/X2 = Kq2/(X+0.5)2
q1/X2 = q2/(X+0.5)2dividing 10-7 both side,
(+2 x 10-7)/X2 = (5 x 10-7)/(X+0.5)2
2/X2 = 5/(X+0.5)2After using quadratic equation, values of X will be:
2(X+0.5)2 = 5X2
2(X2+X+0.25) = 5X2
5X2 = 2X2+2X+0.5
3X2-2X-0.5 = 0
X=0.86, X=-0.19
We neglect the negative value, So,
X=0.86m from q1 (Ans)
