Top 50 Most Repeated ELECTRIC CHARGES AND FIELDS PYQs | JEE MAINS
A curated collection of the most important questions from ELECTRIC CHARGES AND FIELDS, fully solved with step-by-step concepts to prepare for JEE MAINS.
A curated collection of the most important questions from ELECTRIC CHARGES AND FIELDS, fully solved with step-by-step concepts to prepare for JEE MAINS.
The electric field is the negative gradient of the electric potential: $\vec{E} = -\nabla V$. This means the field points in the direction of decreasi...
Read Full Step-by-Step Solution →Conductors induce opposite charges on the side closer to an external charge....
Read Full Step-by-Step Solution →Outside sphere, $E = \frac{1}{4\pi\varepsilon_0} \frac{Q}{r^2}$. At $r=2R$, $E = \frac{Q}{16\pi\varepsilon_0 R^2}$...
Read Full Step-by-Step Solution →System becomes series of conducting slab (infinite $K$) and dielectric; effective capacitance increases by factor $\frac{8}{3}$....
Read Full Step-by-Step Solution →The distance $r$ from the origin to $P$ is $\sqrt{3^2+4^2}=5\,\text{m}$. Using $V = kQ/r$, we get $V = (9\times10^{9})(8\times10^{-6})/5 = 1.44\times1...
Read Full Step-by-Step Solution →Coulomb's law: $F = k\frac{|q_{1}q_{2}|}{r^{2}}$. Substituting $|q_{1}q_{2}| = 3\times10^{-6}\times5\times10^{-6}=15\times10^{-12}\,\text{C}^{2}$ and ...
Read Full Step-by-Step Solution →Induced charges cancel; no net force on conductor in uniform field....
Read Full Step-by-Step Solution →Equate fields: $kq/x^2 = k(4q)/(r-x)^2$. Solving gives $x = r/3$....
Read Full Step-by-Step Solution →Energy $U = \frac{Q^2}{2C}$, $C$ increases by $K$, so $U$ decreases by $K$....
Read Full Step-by-Step Solution →Pairwise potential energies cancel: two positive-negative pairs contribute $-kq^2/a$, one positive-positive contributes $+kq^2/a$, net $U=0$....
Read Full Step-by-Step Solution →For series capacitors,
With battery disconnected, $U \propto 1/K$. So energy becomes $U/4$ when $K=4$....
Read Full Step-by-Step Solution →By Gauss's law, total flux through the cube is $\frac{q}{\epsilon_0}$. Since the cube has 6 identical faces and symmetry ensures equal flux through ea...
Read Full Step-by-Step Solution →$kq/(d/n) - 4kq/(d-d/n) = 0 \Rightarrow n = 4(n-1) \Rightarrow n=5$....
Read Full Step-by-Step Solution →$C \propto A/d$. Doubling $A$ and halving $d$ gives $4\times$ increase....
Read Full Step-by-Step Solution →Induced charges appear; field inside conductor is zero; net force may not be zero due to attraction....
Read Full Step-by-Step Solution →Net potential energy is sum of pairwise interactions; two positive-negative pairs contribute negatively, one positive-positive pair positively....
Read Full Step-by-Step Solution →Both expressions are equivalent due to negative sign. Energy is sum over all pairs....
Read Full Step-by-Step Solution →Each charge contributes
Equal repulsion from two $+q$ charges → resultant along angle bisector toward midpoint....
Read Full Step-by-Step Solution →Since $V$ decreases linearly, $dV/dr$ is negative constant; thus $E = -$(negative) = positive constant....
Read Full Step-by-Step Solution →For points on the axial line far from the dipole, the field magnitude is \(E = \frac{1}{4\pi\varepsilon_0}\frac{2p}{r^{3}}\). The \(r^{-3}\) dependenc...
Read Full Step-by-Step Solution →Distance from P to each charge is 0.10 m. V₁ = k·(+5 μC)/0.10 m = 4.5×10⁵ V, V₂ = k·(–3 μC)/0.10 m = –2.7×10⁵ V. Superposition gives V = V₁ + V₂ = 1.8...
Read Full Step-by-Step Solution →Equating fields $kq/x^2 = k(4q)/(3a-x)^2$ gives $x=a$....
Read Full Step-by-Step Solution →The electric field inside a hollow spherical conductor is zero.\n\text{Step 1: Identify the type of problem.}\n\text{Step 2: Recall the relevant conce...
Read Full Step-by-Step Solution →The electric field is the negative gradient of the electric potential: $\vec{E} = -\nabla V$. This means the field points in the direction of decreasi...
Read Full Step-by-Step Solution →Conductors induce opposite charges on the side closer to an external charge....
Read Full Step-by-Step Solution →Outside sphere, $E = \frac{1}{4\pi\varepsilon_0} \frac{Q}{r^2}$. At $r=2R$, $E = \frac{Q}{16\pi\varepsilon_0 R^2}$...
Read Full Step-by-Step Solution →System becomes series of conducting slab (infinite $K$) and dielectric; effective capacitance increases by factor $\frac{8}{3}$....
Read Full Step-by-Step Solution →The distance $r$ from the origin to $P$ is $\sqrt{3^2+4^2}=5\,\text{m}$. Using $V = kQ/r$, we get $V = (9\times10^{9})(8\times10^{-6})/5 = 1.44\times1...
Read Full Step-by-Step Solution →Coulomb's law: $F = k\frac{|q_{1}q_{2}|}{r^{2}}$. Substituting $|q_{1}q_{2}| = 3\times10^{-6}\times5\times10^{-6}=15\times10^{-12}\,\text{C}^{2}$ and ...
Read Full Step-by-Step Solution →Induced charges cancel; no net force on conductor in uniform field....
Read Full Step-by-Step Solution →Equate fields: $kq/x^2 = k(4q)/(r-x)^2$. Solving gives $x = r/3$....
Read Full Step-by-Step Solution →Energy $U = \frac{Q^2}{2C}$, $C$ increases by $K$, so $U$ decreases by $K$....
Read Full Step-by-Step Solution →Pairwise potential energies cancel: two positive-negative pairs contribute $-kq^2/a$, one positive-positive contributes $+kq^2/a$, net $U=0$....
Read Full Step-by-Step Solution →For series capacitors,
With battery disconnected, $U \propto 1/K$. So energy becomes $U/4$ when $K=4$....
Read Full Step-by-Step Solution →By Gauss's law, total flux through the cube is $\frac{q}{\epsilon_0}$. Since the cube has 6 identical faces and symmetry ensures equal flux through ea...
Read Full Step-by-Step Solution →$kq/(d/n) - 4kq/(d-d/n) = 0 \Rightarrow n = 4(n-1) \Rightarrow n=5$....
Read Full Step-by-Step Solution →$C \propto A/d$. Doubling $A$ and halving $d$ gives $4\times$ increase....
Read Full Step-by-Step Solution →Induced charges appear; field inside conductor is zero; net force may not be zero due to attraction....
Read Full Step-by-Step Solution →Net potential energy is sum of pairwise interactions; two positive-negative pairs contribute negatively, one positive-positive pair positively....
Read Full Step-by-Step Solution →