$5\frac {N}{C}$
Calculating Electric Field Magnitude
The relationship between electric field magnitude ($E$), the force ($F$) experienced by a charge, and the magnitude of the charge ($q$) is given by the formula:
$F = qE$
We need to find the electric field ($E$). We can rearrange the formula to solve for $E$:
$E = \frac{F}{q}$
Applying the Formula
Given:
- Charge ($q$) = 2 C
- Force ($F$) = 10 N
Substitute these values into the rearranged formula:
$E = \frac{10 \text{ N}}{2 \text{ C}}$
$E = 5 \frac{\text{N}}{\text{C}}$
Therefore, the magnitude of the electric field at that point is $5 \frac{\text{N}}{\text{C}}$.
Important Questions from Electric Fields and Gauss' Law
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 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 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:
- 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:
- 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,
$\left[ {{\rm{Take\;}}\frac{1}{{4{\rm{\pi }}{\epsilon_0}}} = 9 \times {{10}^9}{\rm{N}}{{\rm{m}}^2}{{\rm{C}}^{ - 2}}} \right]$
