Two carbon discs, $$1.0\ g$$ each, are $$1.0\ cm$$ apart have equal and opposite charges. If force of attraction between them is $$10^{-5}N$$, the ratio of excess electrons to the total atoms on the negatively charge disc is $$(N_A=6\times 10^{23})$$

$$10^{-13}:2.4$$

$$2.4\times 10^{-12}:1$$

$$10^{12}:2.4$$

$$2.4:10^{12}$$

Open in App

Solution

Verified by Toppr

Correct option is B. $$10^{-13}:2.4$$

Was this answer helpful?

Similar Questions

Two carbon discs, $$1.0\ g$$ each, are $$1.0\ cm$$ apart have equal and opposite charges. If force of attraction between them is $$10^{-5}N$$, the ratio of excess electrons to the total atoms on the negatively charge disc is $$(N_A=6\times 10^{23})$$

View Solution

Two carbon discs of

1.0 g

each are

1.0 c m

apart have equal and opposite charges. If forces of attraction between them is

1.00 \times 10^{- 5} N

, calculate the ratio of excess electrons to total atoms on the negatively charged disc. (permitivity constant is

9.0 \times 10^{9}

N m^{2}

C^{- 2}

)

View Solution

Two carbon discs of 1.0 g each are 1.0 cm apart have equal and opposite charges. If forces of attraction between them is $$1.00 \times 10^{-5}$$N, calculate the ratio of excess electron to total atoms on the negatively charged disc. (Permitivity constant is $$0.0 \times 10^9 N m^2 C^{-2}$$

View Solution

2 equal and opposite charges are placed at a certain distance apart and force of attraction between them is F . If 75% charge of one is transfarred to another, then the force between the charges becomes?

View Solution

Calculate the force of attraction between an electron ans a body having two proton charge when they are

0.529 \times 10^{- 8} c m

apart. Assume charge of one electron and one proton equal to

- 1.6 \times 10^{- 19} C

and

+ 1.6 \times 10^{- 19} C

respectively.

View Solution