Explanation
Hint:-Use the formula of Coulamb's Force in a medium
Explanation:-
∙Force of attraction between two charges in air is given by Fa=q1q24πε0r2
∙When the air is replaced by dielectric constant K, the force of attraction between two charges becomes Fm=q1q24πKε0=FaK
Thus the force is decreasing by K times
Hence, the correct option is (C)
Potential energy =14πϵoq.Qr
|ΔP|=14πϵoq.Qxdx−14πϵoqQx+dx
limdx→0=14πϵo.Q.2xx2−(dx)2
Where dx=0
so the answer is x2
Three charge +4q, Q and q are placed in a straight line of length l at points distance 0, l2 and l respectively. What should be the value of Q in order to make the net force on q to be zero?
A solid insulating sphere of radius a carries a net positive charge 3Q, uniformly Distributed throughout its volume. Concentric with this sphere is a conducting Spherical shell with inner radius b and outer radius c and having a net charge Q, As shown in the figure.Electric field varies with distance r from the centre as (K=14πεo)
A point charge Q is placed at origin O. Let →EA,→EB and →EC represent electric fields at A, B and C respectively. If coordination of A,B and C are respectively (1,2,3) m,(1,1,-1) m and (2,2,2) m then
The chasrge density of an insulating infinity surface is (e/π)cm2 then the field intensity at a nearby point in volt/meter will be -
1.The electric lines of force are imaginary lines the direction of filed lines for the positive charge is away from the centre and for the negative charge the field lines move towards the centre.
2. The field lines can never intersect each other because at the intersection point there can be two direction of field which is not possible.
3. The field originated from positive charge and terminate at the negative charge.
4. The lines of the force in free space is given by E=q4πε0r2
Given that,
Electric field E=500V/m
Potential V=3000V
We know that,
E=Vd
d=VE
d=3000500
d=6m
Hence, the distance is 6 m
Electric field for infinitely long wire is given by E=λ2πε0rwhereλ=chargedensity
Since the electron is revolving then the centripetal force is balanced by the electrostatic force
Fe=Fc
eE=mv2r
eλ2πε0=mv2
eλ4πε0=12mv2
KE=eλ4πε0
Given,
Electric field, E=20NC−1
Area, A=10×10−4m2 A
Electric Flux ϕ=E.Acosθ=20×10×10−4cos60o=0.01Vm
Charge q=10−7C
The electric flux is
ϕ=qε0
ϕ=10−78.85×10−12
ϕ=0.1129×105
ϕ=1.13×104Nm2/C
Hence, the flux is 1.13×104Nm2/C
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