Rankers Physics

Gauss's Law: Practice Problem & Solution

Two large thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite signs and magnitude 27\times 10^{-22}Cm^{-2}. The electric field \overrightarrow{E} in region II in between the plates is : Gauss's Law diagram: Two large thin metal plates are parallel and
\[ 4.25\times 10^{-8}NC^{-1}\]
\[ 6.28\times 10^{-10}NC^{-1}\]
\[ 3.05\times 10^{-10}NC^{-1}\]
\[ 5.03\times 10^{-10}NC^{-1}\]

Short Solution (Best for NEET):

To solve this problem, we apply the core principles of Gauss's Law. Understanding the underlying formula is key to arriving at the correct answer below:

Electric field at Point II will be due to both the sheets so,

\[\overrightarrow{E}=\frac{\sigma}{2\varepsilon_{0}} + \frac{\sigma}{2\varepsilon_{0}}=\frac{\sigma}{\varepsilon_{0}}\]

Leave a Reply

Your email address will not be published. Required fields are marked *

Similar Practice Questions

Explore conceptually related problems to test your knowledge.

1. A dipole of dipole moment $\vec{p}$ is placed in uniform electric field $\vec{E}$ then torque acting on it is given by: (2001)
2. A thin spherical shell of radius R has charge Q spread uniformly over its surface. Which of the following graphs most closely represents the electric field E (r) produced by the shell in the range 0 ≤ r < ∞, where r is the distance from the centre of the shell?
3. The electric field in a certain region is acting radially outward and is given by $E = Ar$. A charge contained in a sphere of radius '$a$' centered at the origin of the field, will be given by: (2015)
4. Electric field at the equator of a dipole is $E$. If strength and distance is now doubled then the electric field will be: (1998)
5. Two parallel plates of infinite dimensions are uniformly charged. The surface charge density on one is \(\sigma_{A} \)and on the other is \(\sigma_{B}\), field intensity at point C will be