Understanding Gauss's Law for Charge Calculation
This problem involves finding the total electric charge enclosed within a cube when given the net electric flux passing through it and the permittivity of free space. The fundamental principle connecting these quantities is Gauss's Law.
Gauss's Law Explained
Gauss's Law states that the net electric flux ($\Phi_E$) through any closed surface is directly proportional to the enclosed electric charge ($Q$) and inversely proportional to the permittivity of free space ($\epsilon_0$). The formula is:
$ \Phi_E = \frac{Q}{\epsilon_0} $
This law is a cornerstone of electrostatics, relating electric fields to their sources (charges).
Applying the Formula
We are given:
- Net flux through the cube, $\Phi_E = 1.05 \text{ N m}^2 \text{ C}^{-1}$
- Permittivity of free space, $\epsilon_0 = 8.85 \times 10^{-12} \text{ C}^2 \text{ N}^{-1} \text{ m}^{-2}$
Our goal is to find the total charge ($Q$) inside the cube. We can rearrange Gauss's Law to solve for $Q$:
$ Q = \Phi_E \times \epsilon_0 $
Calculation Steps
Now, substitute the given values into the rearranged formula:
$ Q = (1.05 \text{ N m}^2 \text{ C}^{-1}) \times (8.85 \times 10^{-12} \text{ C}^2 \text{ N}^{-1} \text{ m}^{-2}) $
Multiply the numerical values:
$ Q = (1.05 \times 8.85) \times 10^{-12} \text{ C} $
$ Q = 9.2925 \times 10^{-12} \text{ C} $
Final Result
Rounding the result to a similar precision as the given values, the total charge inside the cube is approximately $9.29 \times 10^{-12}$ C. This matches one of the options provided.
Important Questions from Electric Fields and Gauss' Law
- Two point charges $q_1 \left( {\sqrt {10} {\rm{\mu C}}} \right)$ and $q_2(-18\sqrt{2} {\rm{\mu C}})$ are placed on the x-axis at $x = 0$ m and $x = 4$ m respectively. The electric field (in V/m) at a point $(1, 3)$ m is,
$\left[ {{\rm{Take\;}}\frac{1}{{4{\rm{\pi }}{\epsilon_0}}} = 9 \times {{10}^9}{\rm{N}}{{\rm{m}}^2}{{\rm{C}}^{ - 2}}} \right]$ - Let a total charge $2Q$ be distributed in a sphere of radius $R$, with the charge density given by $\rho(r) = Cr^2$, where $r$ is the distance from the centre. Two charges $A$ and $B$, of $-Q$ each, are placed on diametrically opposite points, at equal distance, '$a$' from the centre. If $A$ and $B$ do not experience any force, then:
The expression for torque '\(\vec{\tau}\)' experienced by an electric dipole of dipole moment '\(\vec{P}\)' in an external uniform electric field '\(\vec{E}\)' is given by :
The surface charge density of a thin spherical shell placed in an air medium is 88.54 c/m2 The intensity of the electric field measured 12 mm outside the shell from the centre of the shell is 5.625 × 101 2 N/C. The thin spherical shell has a radius of:
The electric flux passing through a surface of area A = 8j m2 in an electric field vector E = 2i + 3j - 4k V/m (bold is for vectors) is:
