Coulomb's Law Calculator

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Coulomb’s law is stated as :

“The magnitude of the electrostatic force between the two point electric charge is directly proportional to the product of the magnitudes of each of the charges and inversely proportional to the square of the distance between the two charges”.

It is given as:

**Step 1 :**

**Step 2 :**

The Coulomb's law Calculator helps to calculate the electrostatic force acting between any two charges or we can even determine the other charge if we know the electrostatic force and any of the two charges.

Coulomb’s law is stated as :

“The magnitude of the electrostatic force between the two point electric charge is directly proportional to the product of the magnitudes of each of the charges and inversely proportional to the square of the distance between the two charges”.

It is given as:

F = $\frac{k q_{1} q_{2}}{d^{2}}$

Where k is Electrostatic constant which is $\frac{1}{4 \pi \epsilon}$

= 9 $\times$ 10^9 N m^{2}/c^{2}

F = Electrostatic Force,

q_{1} = Charge of first body,

q_{2} = Charge of the second body,

d = Distance between two bodies.

Where k is Electrostatic constant which is $\frac{1}{4 \pi \epsilon}$

= 9 $\times$ 10^9 N m

F = Electrostatic Force,

q

q

d = Distance between two bodies.

Analyze the problem and get to know the given parameter

To determine the unknown parameter use the formula

F = K $\frac{q_{1} q_{2}}{d^{2}}$

Where k is Electrostatic constant which is $\frac{1}{4 \pi \epsilon}$

= 9 $\times$ 10^{9} N m^{2}/c^{2}

F = Electrostatic Force,

q_{1} = Charge of first body,

q_{2} = Charge of the second body,

d = Distance between two bodies.

Substitue the given quantities in this formula, calculate and get the unknown quantity.

A charge of q

_{1}=-1.0 $\mu$ C is placed and a second charge, q_{2 }= -10 mC is placed at a distance of 50 cm from the origin. Find the force on q_{1}due to q_{2}, if they are in free space?**Step 1 :**q

_{1}=-1.0 $\mu$ C and

q

_{2}=-10 mC

d = 50 cm = 0.5 m

**Step 2 :**By Coulombs law,

F = $\frac{k q_{1} q_{2}}{d^{2}}$

F = 9 $\times$ 10^{9}$\frac{ -1 \times 10^{-6}}{ -10 \times 10^{-3}}$**Answer :**Electrostatic Force F = - 360 Newton.

A particle is having a charge of 50 C and is placed at a distance of 50 cm having a force of 8 $\times$ 10

^{14}N. Calculate the other Charge?**Step 1 :**q

_{1}= 50 C,

q

_{2}= ?,

d = 50 cm,

F = 8 $\times$ 10

^{14}N**Step 2 :**F = $\frac{k q_{1} q_{2}}{d^{2}}$

q

_{2}= $\frac{F d^{2}}{k q_{1}}$

= $\frac{8 \times 10^{14} \times 0.5^{2}}{9 \times 10^{9}}$

**Answer :**q

_{2}= 4.44 $\times$ 10^{-20}C