Question

two very small spheres A and B are charged with +10 and+20 coulomb respectively and separated by a distance of 80cm. the electric field at point on the line joining the centres of two spheres will be zero at almost how much distance from A

1) 20cm
​2) 33cm
3) 45cm
4) 60cm

Answer 1

hope this will help u...

Answer 2

The distance from A is 33 cm for the electric field which is zero on the line joining the centers of the two spheres.

Solution:

The two small spheres with respective charges are at a distance 80 cm from each other. The net electric field will be zero at a point whose distance is calculated as follows,

Given: Charge A = +10

            Charge B = +20

             R = 80 cm

Let x be the distance for which electric field be zero from A and from B it will be 80 – x

As we know, $E=\frac{k q}{r^{2}}, E_{1}-E_{2}=0$

$\begin{array}{l}{E_{1}=E_{2}} \\\\ {\frac{k q A}{r_{A}^{2}}=\frac{k q B}{r_{B}^{2}}} \\ \\{\frac{q A}{r_{A}^{2}}=\frac{q B}{r_{B}^{2}}} \\\\ {\frac{10}{x^{2}}=\frac{20}{(80-x)^{2}}}\end{array}$

$\begin{array}{l}{\sqrt{\frac{10}{x}}=\sqrt{\frac{20}{80-x}}} \\ \\{\sqrt{\frac{10}{x}}=\frac{\sqrt{10} \sqrt{2}}{80-x}} \\ \\{\frac{1}{x}=\frac{\sqrt{2}}{80-x}} \\ \\{80-x=\sqrt{2 x}}\end{array}$

$\begin{aligned} 80 &=\sqrt{2 x}+x \\ 80 &=1.414 x+x \\ 80 &=2.414 x \\ x &=\frac{80}{2.414} \\ \bold{x &=33 \mathrm{cm}} . \end{aligned}$