(a) Two insulated charged copper spheres A and B have their centers separated by a distance of 50 cm. What is the mutual force of electrostatic repulsion if the charge on each is 6.5 $\times 10^{- 7}$ C? The radii of A and B are negligible compared to the distance of separation.

(b) What is the force of repulsion if each sphere is charged double the above amount, and the distance between them is halved?

10 Views | Posted 5 months ago

Asked by Shiksha User

1 Answer

Answered by

alok kumar singh | Contributor-Level 10

3 months ago

1.12 (a) Charge on sphere A, $q_{A}$ = 6.5 $\times 10^{- 7}$ C

Charge on sphere B, $q_{B}$ = 6.5 $\times 10^{- 7}$ C

Distance between the spheres, r = 50 cm = 0.5 m

Force of repulsion between two spheres

F = $\frac{1}{4 π ?_{0}}$ $\times \frac{q_{1} q_{2}}{r^{2}}$ , where $?_{0}$ = Permittivity of free space = 8.854 $\times 10^{- 12}$
$C^{2} N^{- 1}$ $m^{- 2}$ Therefore, F = $\frac{1}{4 \times π \times {8.854 \times 10^{- 12}}_{}}$ $\times \frac{6.5 \times 10^{- 7} \times 6.5 \times 10^{- 7}}{{0.5}^{2}}$ N = 0.0152 N = 1.52 $\times 10^{- 2}$ N

(b) Charge on sphere A, $q_{A}$ = 2 $\times$ 6.5 $\times 10^{- 7}$ C = 1.3 $\times 10^{- 6} C$

Charge on sphere B, $q_{B}$ = 2 $\times$ 6.5 $\times 10^{- 7}$ C = 1.3 $\times 10^{- 6} C$

Distance between the spheres, r =
$\frac{50}{2}$ = 25 cm = 0.25 m

Force of repulsion between two spheres

F = $\frac{1}{4 π ?_{0}}$ $\times \frac{q_{1} q_{2}}{r^{2}}$ , where $?_{0}$ = Permittivity of free space = 8.854 $\times 10^{- 12}$ $C^{2} N^{- 1}$ $m^{- 2}$ Therefore, F =
$\frac{1}{4 \times π \times {8.854 \times 10^{- 12}}_{}}$ $\times \frac{1.3 \times 10^{- 6} \times 1.3 \times 10^{- 6}}{{0.25}^{2}}$ N = 0.243 N

0 Upvotes 0 Downvotes

Similar Questions for you

Which are the main topics covered in the Electric Charges and Fields?

Pallavi Pathak

The following are the topics covered in this chapter: Electric Field and Field Lines, Gauss's Law, Electric Dipole, Conductors and Insulators, and Electric Flux.

In comparison to other Physics Class 12 chapters, is Electric Charges and Fields an easy chapter to read?

Pallavi Pathak

Indeed, it is an easy chapter of Class 12 Physics. In this chapter, you will learn about the foundational concepts like Gauss's law and electric fields.

A point charge of + $12 μ C$ is at a distance 6cm vertically above the centre of a squared of side 12cm as shown in figure. The magnitude of the electric flux through the square will be……………..× 10³ Nm²/C.

Vishal Baghel

Flux through the 6 sides of square (i.e. cube)

$?_{T} = \frac{q_{i n}}{ε_{0}} = \frac{12 \times 1 0^{- 6} C}{ε_{0}}$

Flux through a square

$? = \frac{?_{T}}{6} = \frac{12 \times 1 0^{- 6}}{ε_{0} \times 6} = \frac{2 \times 1 0^{- 6}}{ε_{0}}$

$\Rightarrow ? = 225.98 \times 1 0^{3} N m^{2} / C ? 226 N m^{2} / C$

Is Electric Charges and Fields chapter important for NEET examination?

Pallavi Pathak

In the medical entrance test NEET, this chapter has a moderate weightage. You can expect around 2-3 questions of this chapter that contributes to the 3-5% of the total marks in the Physics section.

The electric field inside a sphere having charge density related to the distance from the centre as r (a is a constant) is: