JEE MAIN - Physics (2016 - 9th April Morning Slot - No. 12)

The potential (in volts) of a charge distribution is given by.

V(z) = 30 $$-$$ 5x2 for $$\left| z \right|$$ $$ \le $$ 1 m.
V(z) = 35 $$-$$ 10 $$\left| z \right|$$ for $$\left| z \right|$$ $$ \ge $$1 m.

V(z) does not depend on x and y. If this potential is generated by a constant charge per unit volume $${\rho _0}$$ (in units of $${\varepsilon _0}$$) which is spread over a certain region, then choose the correct statement.
$${\rho _0}$$ = 10 $${\varepsilon _0}$$ for $$\left| z \right|$$ $$ \le $$ 1 m and $${\rho _0} = 0$$ elsewhere
$${\rho _0}$$ = 20 $${\varepsilon _0}$$ in the entire region
$${\rho _0}$$ = 40 $${\varepsilon _0}$$ in the entire region
$${\rho _0}$$ = 20 $${\varepsilon _0}$$ for $$\left| z \right|$$ $$ \le $$ 1 m and $${\rho _0} = 0$$ elsewhere

Explanation

We know,

E(z) = $$-$$ $${{dv} \over {dz}}$$

$$ \therefore $$   E(z) = $$-$$ 10 z for $$\left| z \right| \le 1$$ m

and E(z) = 10 for $$\left| z \right| \ge 1$$ m

$$ \therefore $$   The source is an infinity large non conducting thick of thickness z = 2 m.

$$ \therefore $$   E = $${\sigma \over {2{\varepsilon _0}}}$$ = $${{\rho \left( 2 \right)} \over {2{\varepsilon _0}}}$$ = $${\rho \over {{\varepsilon _0}}}$$

$$ \therefore $$   $${\rho \over {{\varepsilon _0}}}$$ = 10

$$ \Rightarrow $$   $$\rho $$ = $${10\,{\varepsilon _0}}$$

Comments (0)

Advertisement