Question

Find the value of electric field along with its direction that would completely balance the weight of the helium nucleus.

Answer 1

Weight of helium nucleus $=6.64\times {10}^{-27}kg=4u$

Charge of helium nucleus $=2\times 1.6\times {10}^{-19}C$

As, $E\frac{F}{q}$

For balancing weight of an helium equilibrium must be maintained between weight and the force.

So, $m\times g=F$

$\therefore m\times g=E\times q$

$E=\frac{m\times g}{q}$

$=\frac{6.64\times {10}^{-27}\times 10}{2\times 1.6\times {10}^{-19}}$

$=2.075\times {10}^{-7}N/C$

Therefore $2.075\times {10}^{-7}N/C$ electric field is required to balance helium nucleus.