0
You visited us 0 times! Enjoying our articles? Unlock Full Access!
Question

Two charges $$3\times 10^{-5}C$$ and $$5\times 10^{4}C$$ are placed at a distance $$10\ cm$$ form each other. Find the value of electrostatic force acting between them.

A
$$180\times 10^{9}N$$
B
$$40\times 10^{11}N$$
C
$$13.5\times 10^{11}N$$
D
$$13.5\times 10^{10}N$$
Solution
Verified by Toppr

Correct option is A. $$13.5\times 10^{11}N$$
$$q_{1}=3\times 10^{-5}$$ $$q_{2}=5\times 10^{4}$$
$$r=10\times 10^{-2}m$$
$$F_{e}=\dfrac{1}{4\pi \varepsilon_{0}}\dfrac{q_{1}q_{2}}{r^{2}}$$
$$\dfrac{9\times 10^{9}\times 3\times 10^{-5}\times 5\times 10^{4}}{(10\times 10^{-2})^{2}}$$
$$\dfrac{9\times 15\times 10}{100\times 10^{-4}}$$
$$\dfrac{135}{100}\times 10^{12}$$
$$=1.35\times 10^{12}$$
$$=13.5\times 10^{11}$$

Was this answer helpful?
1
Similar Questions
Q1
Two charges $$3\times 10^{-5}C$$ and $$5\times 10^{4}C$$ are placed at a distance $$10\ cm$$ form each other. Find the value of electrostatic force acting between them.
View Solution
Q2
Two charged particles are placed at a distance of 1.0 cm apart. What is the minimum possible magnitude of electrostatic force acting on each charge
View Solution
Q3
The electrostatic force between charges of 200μC and 500μCplaced in free space is 5×107N. Find the distance between the two charges.
View Solution
Q4
Two point charges q1 and q2 are placed at a distance of 50 m from each other in air, interact with a certain electrostatic force. If the medium is replaced with oil whose relative permittivity is 5, then the interaction force between them is still the same. Now the separation between them is
View Solution
Q5
An attractive force of 9 N acts between +5 μC and 5 μC at some distance. These charges are allowed to touch each other and are then again placed at their initial position. The force acting between them will be :
View Solution