Two point particles a and b having charges of 4*10^-6 and -64×10^-6 respectively are held at a separation of 90 cm locate the points on the lineab or on its extension where the electric field is 0
Answers
Answer:
The net field will be zero at a point at a distance of 6cm from the positive charge and at a distance of 96cm from the negative charge.
Explanation:
The electric field due to a positive charge will be directed away from it, while that due to a negative charge will be directed towards it.
The electric field will be zero at a point nearer to positive charge at some distance d and the distance due to negative charge will be equal to he distance d + distance between the two charges.
The field due to a charge q at a distance d from it is given by the formula
E=k×q÷d
The net field due to both the charges will be zero at a point
(k×4×10^-6)÷d=(k×64×10^-6)÷(90+d)
90+d=16d
15d=90
d=6cm
The electric field cannot be zero in regions II and III
It can be zero only in regions 'I'.
(i) Let electric field is zero at point 'P' in I at distance 'x' from point A.
Equating electric fields at point P, due to both charges [x in cm]
k (4*10^6) / x^2 = k(64*10^6)/(90+x)^2
[ (90 + x) / x ] ^2 = 16
(90+x)/x = 4
x = 30
Hence, At distance x = 30 cm from A along BA.
