Question

two point charges of +3 and -3microcoulomb are placed 2*10^-3m apart from each other. calculate (I) electric field and electric potential at a distance of 0.6m from the dipole in broad side on position (II) electric field and electric potential at the same point after rotating the dipole through 90degree

Answer 1

Answer:

Explanation:

We know that

Electrostatic Potential and Electric field due to dipole, p at any point (r, θ) is given by

V=14πε0p cosθr2, andE=14πε0pr33 cos2θ+1−−−−−−−−−√

Here, dipole moment is

p=ql=3×10−6 C×2×10−3 m=6×10−9 Cm

r = 0.6 m

(i) Here the point is on the axis perpendicular to line joining two charges (dipole) , θ=900⇒cos θ=0

Therefore,

V = 0

E=14πε0pr3=9×109×6×10−90.63N/C=250 N/C

(ii) When dipole is rotated to 900 the point will shift to the axis of dipole, in this case θ=00⇒cosθ=1

Therefore,

V=14πε0pr2=9×109×6×10−90.62V=150 V

E=14πε02pr3=9×109×2×6×10−90.63N/C=500 N/C