JEE MAIN · PHYSICS · ELECTROSTATICS
JEE Main Electrostatics Questions & Solutions
15 solved questions on Electrostatics, ranging from easy to JEE-Advanced-flavour hard. Click any to see the full solution.
15 solved questions on Electrostatics, ranging from easy to JEE-Advanced-flavour hard. Click any to see the full solution.
When a dielectric is inserted into an isolated (disconnected) charged capacitor, the voltage across the plates:
View solution →The dielectric constant of vacuum is exactly 1.
View solution →Introducing a dielectric slab (dielectric constant $K$) fully into a parallel-plate capacitor increases its capacitance by a factor of:
View solution →A 10 μF capacitor is charged to 100 V. Energy stored is $x$ mJ. Find $x$.
View solution →Two capacitors $C_1 = 2\,\mu F$ and $C_2 = 3\,\mu F$ are connected in parallel. The equivalent capacitance is:
View solution →The capacitance of a parallel-plate capacitor doubles if:
View solution →For a charged conducting sphere of radius $R$, the field inside (at $r<R$) is:
View solution →The electric field at distance $r$ from a long uniformly charged wire (linear charge density $\lambda$) is:
View solution →A point charge $Q$ is placed at the center of a cube. The flux through one face is:
View solution →Inside a conductor in electrostatic equilibrium, the electric field is zero.
View solution →A charge of $+4\,\mu C$ is placed at origin. The electric field at a point 2 m away is:
View solution →Electric field lines:
View solution →Two charges $+q$ and $-q$ placed 2 m apart attract with 9 N. Find $q^2/10^{-9}$ in C² (take $k=9\times 10^9$).
View solution →If the distance between two charges is doubled, the electrostatic force becomes:
View solution →Two point charges of $+2\,\mu C$ each are separated by 1 m in vacuum. The force between them is (take $k = 9\times 10^9$ N·m²/C²):
View solution →PrepWiser turns these solved questions into a daily practice loop. Chapter-wise drills, full mocks, AI doubt chat. No auto-renew.
Start free →