Question

The sphere of charge 6.24 uC distributed uniformly throughout its volume. The radius of
the sphere is 2.0 cm. Find the electric field at distances 1.0 cm, 2.0 cm, and 3.0 cm.
(a) 7.02 x107 N/C, 14.0x107 N/C, and 6.24 x107 N/C
(b) 9.31x107N/C, 1.25x107 N/C, and 7.14x107 N/C
(c) 2.33x107 N/C, 12.0x107 N/C, and 23.1x107N/C
(d) 4.31x107N/C, 25.1*107 N/C, and 25.7x107 N/C​

Answer 1

The correct option is A  i.e. 7.02 x10⁷ N/C, 14.0x10⁷ N/C, and 6.24 x10⁷ N/C

GIVEN -

charge of sphere = 6.24ùC

radius of sphere = 2.0 cm

To find - the electric field at distances 1.0 cm, 2.0 cm, and 3.0 cm

SOLUTION

as  F= q.E where f is force , q is charge and E is the electric field

Hence $E=\frac{F}{q}$  

E = $\frac{Kq}{r^{2} }$  where K is constant

For a distance of 2 cm

Firstly we need to convert cm into m i.e. 0.02m

                            E =  $\frac{9X10^{9} X 6.24 X 10^{-6} }{0.02^{2} }$

                             E = $\frac{56.16 X 10^{3} X10^{4} }{0.02X0.02}$

                         E = 14.04 X 10⁷ N/C

For a distance of 1 cm

In this we can just half the electric field

                                                           E =     $\frac{14.04X10^{7} }{2}$

                                                           E = 7.02 X 10⁷ N/C

For a distance of 3 cm

                             $E = \frac{56.16 X 10^{3}X 10^{4} }{0.03X0.03}\\\\ E = 6.24 X 10^{7 } N/C$

Hence,

the electric field for a distance of 1 cm ,2cm and 3 cm are 7.02 x10⁷ N/C, 14.0x10⁷ N/C, and 6.24 x10⁷ N/C