A sphere of charge of radius R carries a positive charge whose volume charge density depends only on the distance…..

Q. A sphere of charge of radius R carries a positive charge whose volume charge density depends only on the distance r from the ball’s centre as ρ = ρ0 (1 − r/R). Where ρ0 is a constant. Assume ε  as the permittivity of space. The magnitude of electric field as a function of the distance r inside the sphere is given by

(a) \displaystyle \frac{\rho_0}{\epsilon}[\frac{r}{3}-\frac{r^2}{4R}]

(b) \displaystyle \frac{\rho_0}{\epsilon}[\frac{r}{4}-\frac{r^2}{3R}]

(c) \displaystyle \frac{\rho_0}{\epsilon}[\frac{r}{3}+ \frac{r^2}{4R}]

(d) \displaystyle \frac{\rho_0}{\epsilon}[\frac{r}{4}-\frac{r^2}{3R}]

Ans: (a)

Sol: charge inside a sphere of radius r (r < R)

\displaystyle q = \int_{0}^{r} \rho dv

\displaystyle = \int_{0}^{r} \rho_0 (1-\frac{r}{R})4\pi r^2 dr

\displaystyle q = 4\pi \rho_0 \int_{0}^{r}(r^2 -\frac{r^3}{R})

\displaystyle q = 4\pi \rho_0 (\frac{r^3}{3}-\frac{r^4}{4R})

\displaystyle E = \frac{1}{4\pi \epsilon}\frac{q}{r^2}

\displaystyle E = \frac{1}{4\pi \epsilon}.4\pi \rho_0 (\frac{r}{3}-\frac{r^2}{4R})

\displaystyle \frac{\rho_0}{\epsilon}(\frac{r}{3}-\frac{r^2}{4R})

Potential difference between centre and surface of the sphere of radius R and uniform volume charge density ρ within it will be

Q. Potential difference between centre and surface of the sphere of radius R and uniform volume charge density ρ within it will be

(a) \displaystyle \frac{\rho R^2}{6\epsilon_0}

(b) \displaystyle \frac{\rho R^2}{4\epsilon_0}

(c) \displaystyle \frac{\rho R^2}{3\epsilon_0}

(d) \displaystyle \frac{\rho R^2}{2\epsilon_0}

Ans: (a)
Charge on Sphere is

\displaystyle q =  \frac{4}{3}\pi R^3 \rho

Electric Potential at the centre

\displaystyle V_o = \frac{1}{4\pi \epsilon_0}\frac{3q}{2R}

Electric Potential at the Surface

\displaystyle V_P = \frac{1}{4\pi \epsilon_0}\frac{q}{R}

Potential difference between centre and surface

\displaystyle  = \frac{1}{4\pi \epsilon_0}\frac{3q}{2R} - \frac{1}{4\pi \epsilon_0}\frac{q}{R}

\displaystyle  = \frac{1}{4\pi \epsilon_0}\frac{q}{2R}

\displaystyle  = \frac{1}{4\pi \epsilon_0 (2R)}(\frac{4}{3}\pi R^3 \rho)

\displaystyle = \frac{\rho R^2}{6\epsilon_0}

Six charges are placed at the vertices of a rectangular hexagon as shown in the figure. The electric field on the line…..

Q. Six charges are placed at the vertices of a rectangular hexagon as shown in the figure. The electric field on the line passing through point O and perpendicular to the plane of the figure as a function of distance x from point O is (x >> a)

Numerical

(a)Zero

(b) \displaystyle \frac{Qa}{\pi\epsilon_0 x^3 }

(c) \displaystyle \frac{2Qa}{\pi\epsilon_0 x^3 }

(d) \displaystyle \frac{\sqrt3 Qa}{\pi\epsilon_0 x^3 }

Ans: (b)

Electric Field due to dipole at a point on the axis passing through O is

\displaystyle E = \frac{p}{4\pi \epsilon_0 x^3}

Where p = Q × 2a

Net field is
\displaystyle E_{net} = 2E = \frac{2 \times Q \times 2 a}{4\pi \epsilon_0 x^3}

\displaystyle = \frac{Q a}{\pi\epsilon_0 x^3 }

Two thin wire rings each having radius R are placed at a distance d apart with their axes coinciding…..

Q. Two thin wire rings each having radius R are placed at a distance d apart with their axes coinciding. The charges on the two rings are +Q and -Q The potential difference between the centres of the two rings is

(a)Zero

(b) \displaystyle \frac{Q}{4\pi\epsilon_0 } [\frac{1}{R}-\frac{1}{\sqrt{ R^2 +d^2}}]

(c) \displaystyle \frac{Q}{4\pi\epsilon_0 d^2}

(d) \displaystyle \frac{Q}{2\pi\epsilon_0 } [\frac{1}{R}-\frac{1}{\sqrt{ R^2 +d^2}}]

Ans: (d)

Sol:
\displaystyle V_1 = \frac{-Q}{4\pi \epsilon_0 \sqrt{R^2 + d^2}} + \frac{Q}{4\pi \epsilon_0 R}

\displaystyle V_2 = \frac{Q}{4\pi \epsilon_0 \sqrt{R^2 + d^2}} + \frac{-Q}{4\pi \epsilon_0 R}

\displaystyle V_1 -V_2 = \frac{Q}{2\pi\epsilon_0 } [\frac{1}{R}-\frac{1}{\sqrt{ R^2 +d^2}}]