The electric field is a vector space property surrounding a source charge that exerts a force on any other test charge placed within it. It is defined quantitatively as the electrostatic force experienced per unit positive test charge.
Calculation of electric field due to point charges, discrete charge distributions, linear, surface, and volume charge distributions, and basic superposition principle.
Continuous charge distributions, Gauss's law applications for symmetric charge systems (spherical, cylindrical, planar), and differential form of Gauss's law.
Conditions: A stationary point charge q in vacuum or free space.
Result: $E = \frac{1}{4piepsilon_0} \frac{q}{r^2} \hat{r}$
Conditions: Uniform surface charge density sigma on an infinite, non-conducting plane sheet.
Result: $E = \frac{\sigma}{2epsilon_0}$
| Formula / law | Equation | Variables and units | Conditions |
|---|---|---|---|
| Electric Field Definition | $$E = \lim_{q_0 \to 0} \frac{F}{q_0}$$ | F = Electrostatic force, $q_0 = Test charge$ N/C | Valid generally |
| Electric Field of a Point Charge | $$E = \frac{1}{4piepsilon_0} \frac{q}{r^2}$$ | q = Source charge, r = Distance N/C | In vacuum or air |
| Electric Field of an Infinite Line Charge | $$E = \frac{\lambda}{2piepsilon_0 r}$$ | lambda = Linear charge density, r = Radial distance N/C | Infinite length approximation |
| Electric Field of an Infinite Plane Sheet | $$E = \frac{\sigma}{2epsilon_0}$$ | sigma = Surface charge density N/C | Non-conducting uniform sheet |
| Electric Field near a Conducting Surface | $$E = \frac{\sigma}{\epsilon_0}$$ | sigma = Surface charge density of conductor N/C | Surface of any charged conductor |
| Electric Field of a Uniform Spherical Shell (r > R) | $$E = \frac{1}{4piepsilon_0} \frac{Q}{r^2}$$ | Q = Total charge, R = Radius of shell, r = Distance from center N/C | Outside the shell |
| Electric Field inside a Uniform Spherical Shell (r < R) | $$E = 0$$ | r = Distance from center N/C | Inside a hollow charged conductor or shell |
| Gauss's Law | $$\oint E \cdot dA = \frac{q_{enclosed}}{\epsilon_0}$$ | E = Electric field, dA = Area vector, $q_enclosed = Net enclosed charge$ N m^2 / C | Any closed hypothetical surface |
Draw: A rectangular hyperbola curve showing E decreasing non-linearly as 1/r^2 with increasing distance r.
Interpret: Demonstrates the inverse-square law relationship between electric field magnitude and distance from a point source.
Exam use: Frequently asked in MCQs to identify correct graphical trends for Coulombic fields versus dipole fields (1/r^3).
Draw: E = 0 from r = 0 up to r = R, followed by a sharp step-up at r = R, then falling off as 1/r^2 for r > R.
Interpret: Reflects the physical principle that static electric field inside a conductor is zero and charge resides entirely on the surface.
Exam use: Tested in conceptual conceptual matching and numerical threshold questions.
Draw: Schematic of an atomizer, viewing microscope, parallel capacitor plates, and ionizing X-ray source.
Interpret: Measures the elementary electronic charge e by balancing gravitational force, buoyant force, viscous drag, and upward electric force (qE = mg).
Exam use: Historical experiment questions regarding charge quantization and electric field balance.
Answer: 9 x 10^5 N/C
Answer: 10^6 N/C
Answer: 3 x 10^4 N m^2/C
Common error: Electric field lines physically exist as tangible threads in space.
Correct: Electric field lines are merely a geometrical visualization tool to represent field direction and relative intensity.
They do not have a physical substance; space between them is continuous.
Common error: A test charge must always be positive and static.
Correct: A test charge is an idealized mathematical point charge assumed to be infinitesimally small so it does not disturb the source charge distribution.
Its sign only determines the conventional direction of the force vector, but the field itself exists independently of the test charge.
| Parameter | Non-Conducting Infinite Plane Sheet | Charged Conducting Surface |
|---|---|---|
| Formula | $E = \frac{\sigma}{2\epsilon_0}$ | $E = \frac{\sigma}{\epsilon_0}$ |
| Charge Distribution | Uniformly distributed throughout or on both faces contributing equally | Entire charge resides on the surface with zero field inside |
| Field Direction | Normal to the sheet, pointing outwards for positive $\sigma$ | Perpendicular to the conducting surface |
| Parameter | Inside Shell ($r < R$) | Outside Shell ($r > R$) |
|---|---|---|
| Electric Field ($E$) | $0$ | $\frac{1}{4\pi\epsilon_0} \frac{Q}{r^2}$ |
| Enclosed Charge ($q_{enclosed}$) | $0$ | $Q$ (Total charge of the shell) |
| Dependency on Distance ($r$) | Independent of $r$ | Inversely proportional to $r^2$ |
Choose an option to reveal the explanation.
Explanation: Electric field intensity is force per unit charge ($E = F/q_0$). Dimensions of force are $[M L T^{-2}]$ and charge is $[A T]$, yielding $[M L T^{-3} A^{-1}]$.
Explanation: Electric field lines are continuous curves that start on positive charges and end on negative charges; electrostatic field lines never form closed loops because electrostatic field is conservative.
Explanation: By Gauss's Law, any Gaussian surface drawn inside a hollow charged conductor encloses zero net charge ($q_{enclosed} = 0$), hence the electric field inside is zero.
Explanation: Applying Gauss's law with a cylindrical Gaussian surface around an infinite line charge yields $E \cdot (2\pi r L) = \frac{\lambda L}{\epsilon_0}$, which simplifies to $E = \frac{\lambda}{2\pi\epsilon_0 r}$.
Explanation: Electric field intensity is measured in both Newtons per Coulomb ($N/C$) from force per charge definition and Volts per meter ($V/m$) from potential gradient.
Explanation: The linear nature of Coulomb's law and Maxwell's equations ensures that the total field is the direct vector sum of individual electric fields.
Explanation: For a non-conducting uniform plane sheet, application of Gauss's law gives $E = \frac{\sigma}{2\epsilon_0}$.
Explanation: If two field lines intersected, the electric field vector at that single point would have two distinct directions, violating the uniqueness of the electric field vector.
Explanation: For points outside a uniform spherical shell ($r > R$), the shell behaves as if all its charge is concentrated at its center, giving $E = \frac{1}{4\pi\epsilon_0} \frac{Q}{r^2}$.
Explanation: Gauss's law states that the surface integral of the electric field over a closed surface equals $\frac{q_{enclosed}}{\epsilon_0}$.
Explanation: A test charge is chosen to be infinitesimally small so that it does not disturb or alter the position of the source charge distribution.
Explanation: For any charged conductor, the electric field right at its surface is directed normally and has a magnitude of $E = \frac{\sigma}{\epsilon_0}$.
Explanation: According to Gauss's Law, the total electric flux through any closed surface enclosing a charge $q$ depends only on the enclosed charge, which is $\Phi = \frac{q}{\epsilon_0}$.
Explanation: The electric field acts as a physical mediator for electrostatic interactions, propagating through space at the speed of light in vacuum.
Explanation: From Coulomb's law, the electric field of a point charge is $E = \frac{1}{4\pi\epsilon_0} \frac{q}{r^2}$.
These are pattern-based practice questions. Authentic year labels are only used when verified from a source.
Explanation: For the net electric field to be zero, the individual fields must oppose each other. This can only happen outside the charges on the side of smaller magnitude, i.e., closer to the $+2\mu C$ charge.
Explanation: The electric field inside any charged conductor (solid or hollow) in electrostatic equilibrium is always zero.
Explanation: Net flux $\Phi_{net} = \Phi_2 - \Phi_1 = \frac{q_{enclosed}}{\epsilon_0}$, which implies $q_{enclosed} = (\Phi_2 - \Phi_1)\epsilon_0$.
Explanation: Using $E = \frac{\lambda}{2\pi\epsilon_0 r} \implies \lambda = E \cdot (2\pi\epsilon_0 r) = 2\pi\epsilon_0 \times (9 \times 10^4) \times (0.02) = \frac{9 \times 10^4 \times 0.04}{9 \times 10^9} = 0.2 \times 10^{-6}\,C/m = 0.2\,\mu C/m$.
Explanation: For equilibrium, downward gravitational force $mg$ must balance upward electrostatic force $qE$, hence $qE = mg \implies E = mg/q$.
Explanation: The electric field at the surface of any charged conductor is always directed perpendicular (normal) to the surface.
Attempt these questions under a ten-minute limit using a separate timer.
Explanation: $E = \frac{1}{4\pi\epsilon_0} \frac{q}{r^2} = \frac{9 \times 10^9 \times 4 \times 10^{-6}}{(0.2)^2} = \frac{3.6 \times 10^4}{0.04} = 9 \times 10^5\,N/C$? Wait, $3.6 \times 10^4 / 0.04 = 9 \times 10^5$, let's check: $9 \times 10^9 \times 4 \times 10^{-6} = 36000$. $0.2^2 = 0.04$. $36000 / 0.04 = 900,000 = 9 \times 10^5$. Correct option is index 0 ($9 \times 10^5\, N/C$).
Explanation: Since $E \propto \frac{1}{r^2}$, doubling the distance ($r \to 2r$) reduces the electric field to $\frac{1}{4}$th of its original value (quartered).
Explanation: For a non-conducting sheet, $E = \frac{\sigma}{2\epsilon_0} = \frac{8.85 \times 10^{-12}}{2 \times 8.854 \times 10^{-12}} \approx 0.5\, N/C$.
Explanation: According to Gauss's law, $\oint E \cdot dA = \frac{q_{enclosed}}{\epsilon_0}$, which depends solely on the net charge enclosed by the Gaussian surface.
Explanation: Magnitude $|E| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5\, N/C$.
Explanation: For an infinite line charge, $E = \frac{\lambda}{2\pi\epsilon_0 r}$, showing inverse proportionality to $r$ ($r^1$). Point charges and spherical shells vary as $1/r^2$, while plane sheet field is uniform (independent of $r$).
Explanation: A corner of a cube is shared by 8 adjacent cubes in a symmetrical lattice. Thus, only $\frac{1}{8}$th of the total flux from the charge passes through the given cube, giving $\frac{q}{8\epsilon_0}$.
Explanation: A positive test charge experiences a force in the exact same direction as the electric field vector ($F = q_0 E$). Thus, it moves along the positive x-direction.
Explanation: In electrostatic equilibrium, excess charge on a conductor resides entirely on its outer surface, making the electric field everywhere inside the solid conductor zero.
Explanation: If there were a tangential component of the electric field at the surface of a conductor, free electrons would experience a force and move, contradicting the condition of electrostatic equilibrium.