electrostat graph...................

yes if u have an arihant book check it out.

@ rahul if it is a conducting sphere then the charges will be repelled and will get arranged on the periphery!

@ Ishan, ques indicates tht
sphere is charged uniformly throughout the volume of the sphere.Then how can electric field at an internal point be 0?

answer is b (taking the sphereto be conducting)

explaination:

we know that the electric field inside a sphere is zero (shell law which is basically to satisfy the electrostat condition) hence the potential will be constant for r'=0 to r'=r (where r is the radius)

now once we r outside the sphere ,the charge can be considered to be concentrated at the centre of the sphere by spherical symmetry hence it will vary inversely with the square of distance from the centre

graph of a constant function is a straight line parallel to x axis (...i.e when r'<=r)

graph of 1/x² is a rectangular hyperbola restricted in the first quadrant

join these two and the answer..i.e (B) becomes evident!!

the ans would had been b if it was a hollow sphere the first line says it is a solid sphere

take k=1/4(pi)e_o
The electric field intensityoutside the sphere is :
E_outside=k q/r²
dV_outsidedr = -E_outside
or
\int_{\propto }^{V}{dV_outside} = - \int_{\propto }^{r}{kq/r²} dr

or V = kq/r as V_\propto = 0

or V \propto 1/r

At r=R V=kq/R

i.e at the surface of the sphere potential is V_s = kq/R

The electric intensity inside the sphere ,
E_inside = k q r/R³

dV_insidedr = - E_inside

or \int_{V_s}^{V}{dV_inside} = -kqR³\int_{R}^{r}{r dr}

so V - V_s = - kqR³[r²2]^r_R

substituting V_s = kqR(32 - r²2R²)

At center r=0 & V_c = (3/2)(kq/R) = 1.5 V_s

i.e. the potential at the center is 1.5 times the potential at the surface

graph will be as my previous post...........

E_outside = \frac{1}{4\pi \varepsilon _{0}} \frac{q}{r^{2}} \frac{dV_{outside}}{dr} = - E_{outside} \int_{infinite}^{V}{dV_{outside}} = - \int_{infinite}^{r}{\frac{1}{4\pi \varepsilon _{0}} \frac{q}{r^{2}} }dr V = \frac{1}{4\pi \varepsilon _{0}} \frac{q}{R} at r = R E_{inside} = \frac{1}{4\pi \varepsilon _{0}} .\frac{q}{R^{3}}.r \frac{dV_{inside}}{dr} = - E_{inside} \int_{V_{s}}^{V}{dV_{inside}}= - \frac{1}{4\pi \varepsilon _{0}}\frac{q}{R^{3}}\int_{R}^{r}{r}dr where V_s is potential at surface = \frac{1}{4\pi \varepsilon _{0}} \frac{q}{R}

V - V_s = - \frac{1}{4\pi \varepsilon _{0}} \frac{q}{R^{3}}[\frac{r^{2}}{2}] with integration limits R to r

Substituting V_s, we get V = \frac{1}{4\pi \varepsilon _{0}} \frac{q}{R}\left[\frac{3}{2} - \frac{1}{2}\frac{r^{2}}{R^{2}}] \right

GOT THAT !!!!!

no i dont hav arihant , plzzzzzz post that derivation

well assuming sphere is positively charged!!![3]
and should be a

saw in arihant ans c

the graph is

So if it is so ans is C)

charged uniformly means dat the sphere is a nonconductor.

c.

is it b)