Question

A spherical cavity made in a informaly changed non-conducting sphere and a positive charge in released from position $P$ as shown in fig. Then path followed by particle is

• A. A straight line along radically upward of cavity
• B. A straight line along radially outward of cavity
• C. A straight line along positive $X-direction$.
• D. A straight line along radially outward of sphere.

Answer 1

The correct option is A A straight line along radically upward of cavity

A spherical cavity made in an informally changed non-conductivity sphere

a positive charge in released from position P

The path followed by particle

using Lorentz Force equation

$\overrightarrow{F_q}=q\overrightarrow{E}+q\overrightarrow{V}\times B$

For a charge at rest $\overrightarrow{V}=0$ so the above equation becomes

$\overrightarrow{F_q}=q\overrightarrow{E}$

This is scalar product of the vector.

This means $\overrightarrow{F_q}$ and $\overrightarrow{E}$ are either parallel or antiparallel to each other hence the path will be a straight line along radially upward of cavity.