Question

two equal point charges each of 3 uC are separated by a certain distance in metres.if they are located at (i+j+k) and (2i+3j+3k) then the electrostatic force between them is

Answer 1

F will be minimum when q1andq2q1andq2 are minimum i.e.q1=q2=1.6×10−19i.e.q1=q2=1.6×10−19

Also, given r=1mr=1m.

Fmin=9×109N×(1.6×10−10)2Fmin=9×109N×(1.6×10−10)2

Fmin=2.3×10−28N.

Answer 2

Concept:

The study of electric charges at rest is electrostatics, a branch of physics. Some materials, such as amber, have been known to attract lightweight particles after rubbing since classical times. The word 'electrical' comes from the Greek word v, which means amber.

Given:

Two points having charge $3uC$ located at $(i+j+k)$ and$(2i+3j+3k)$.

To Find:

The electrostatic force between the points.

Solution:

$r$ is the distance between two charges.

$r=\sqrt{(2-1)^{2}+(3-1)^{2}+(1-1)^{2}}\\\\ r=\sqrt{5}$

The force between two charges, according to Coulomb's law, is F.

$F=\frac{1}{4 \pi \epsilon_{0}} \frac{\mathrm{q}_{1} \mathrm{q}_{2}}{\mathrm{r}^{2}}$

$(9\times \left.10^{9}\right)\left(\frac{3 \times 10^{-6} \times 3 \times 10^{-6}}{5}\right)=16.2 \times 10^{-3} \mathrm{~N}$

The electrostatic force between the points is $16.2 \times 10^{-3} \mathrm{~N}$.

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