Question

two point charges of 3 microcoulomb and -3 microcoulomb are placed 0.002 m apart calculte 1 electric field and electri potential at a distance of 0.6 from the dipole in broad side on position 2 also find field and potential when dipole is rotated by 90 degree​

Answer 1

Answer:

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Answer 2

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