WorksheetsElectrostatics Revision
Total questions: 19
Worksheet time: 20mins
The electric field intensity at any point is the force experienced by unit positive charge, given by
E=F/q
E=qF
E=U/q
E=V.q
Electric field and potential due to a point charge
Kq2/r2, Kq/r
Kq/r2, Kq/r
Kq/r2, Kq2/r
Kq2/r, Kq/r
Electric field and potential due to Infinitely long line charge
E=2Kλ/d, V=-2Kλ*ln(ra-rb)
E=Kλ/d(cos90+sin90), V=2Kλ*ln(ra/rb)
E=2Kλ/3d, V=V=-2Kλ*ln(ra-rb)
E=2Kλ/d, V=-2Kλ*ln(ra/rb)
E and V due to Infinite long sheet
E=σ/2ε, Va-Vb=(σ/2ε)(rb-ra)
E=σ/ε, Va-Vb=(σ/2ε)(rb-ra)
E=σ/2ε, Va-Vb=(σ/2ε)*(ln(ra-rb))
It's easy, use logic
Electric field and Potential due to a ring
E=KQx/(x2+r2)3/2, V=KQ/(x2+r2)1/2
E=KQ/(x2+r2)3/2, V=KQ/R
Ecentre=0, Vcentre=0
E=KQx/(x2+r2)1/2, V=KQx/r
E and V due to Uniformly charged hollow conducting/nonconducting/solid conducting sphere is given by E=KQ/r2, V=KQ/r
Electric field and Potential due to Uniformly charged solid nonconducting sphere
E=ρr/3ε (inside the sphere)
E=Kq/r2, V=Kq/r (outside the sphere)
V=KQ(3R2-r2)/2R3 (inside the sphere)
V=(3/2)*(KQ/R) (at centre)
Electric field due to disc is (σ/2ε)(1-cosθ), where θ is angle between hypotenuse and axial line
Electric field at axial point of short dipole
Electric field at equatorial axis of a short dipole
Electric field at axis of dipole and equatorial axis respectively is 2Kpr/(r2-l2)2 and Kp/(r2+l2)3/2
Electric field die to dipole at general point is E=(Kp/r3)√(1+_tan2θ)
(a)
Potential due to dipole
Kpcosθ∕r2
Kpcosθ∕r3
Potential energy of Dipole in electric field
Torque on Dipole kept in a electric field at angle θ
The relation between E and V is V=-∫E.dl
U is proportional to
V
V2
V3
V-1
Unet=U1+U2+U3+...
Electric potential is a vector quantity and can be added normally
YES
NO
