VITEEE PYQs for Electric charges and fields with Solutions: Practice VITEEE Previous Year Questions

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Shivam Yadav

Updated on - Dec 9, 2025

Electric charges and fields is an important topic in the Physics section in VITEEE exam. Practising this topic will increase your score overall and make your conceptual grip on VITEEE exam stronger.

This article gives you a full set of VITEEE PYQs for Electric charges and fields with explanations for effective preparation. Practice of VITEEE Physics PYQs including Electric charges and fields questions regularly will improve accuracy, speed, and confidence in the VITEEE 2026 exam.

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VITEEE PYQs for Electric charges and fields with Solutions

VITEEE PYQs for Electric charges and fields with Solutions

  • 1.

    The electric potential \( V(x) \) in a region around the origin is given by \( V(x) = 4x^2 \, {volts}. \) .The electric charge enclosed in a cube of 1m side with its center at the origin is (in coulomb)

      • \( 8\epsilon_0 \)
      • \( -4\epsilon_0 \)
      • 0
      • \( -8\epsilon_0 \)

    • 2.
      An object is thrown vertically upwards. At its maximum height, which of the following quantity becomes zero?

        • Momentum
        • Potential energy
        • Acceleration
        • Force

      • 3.
        A solid sphere of radius $R_{1}$ and volume charge density $\rho=\frac{\rho_{0}}{r}$ is enclosed by a hollow sphere of radius $R_{2}$ with negative surface charge density $\sigma$, such that the total charge in the system is zero, $\rho_{0}$ is a positive constant and $r$ is the distance from the centre of the sphere. The ratio $\frac{R_{2}}{R_{1}}$ is

          • $\frac{\sigma}{\rho_{0}}$
          • $\sqrt{2\sigma/ \rho_{0}} $
          • $\sqrt{\rho_{0} /\left(2\sigma\right)} $
          • $\frac{\sigma }{\rho _{0}} $

        • 4.

          An element of \( 0.05 \) m is placed at the origin, carrying a large current of \( 10 A \). The magnetic field at a perpendicular distance of \( 1 \) m is: 
           

            • \( 4.5 \times 10^{-8} \) T
            • \( 5.5 \times 10^{-8} \) T
            • \( 5.0 \times 10^{-8} \) T
            • \( 7.5 \times 10^{-8} \) T

          • 5.
            The potential energy of a particle \( U(x) \) executing simple harmonic motion is given by:

              • \( U(x) = \frac{k}{2} (x - a)^2 \)
              • \( U(x) = k_1 x + k_2 x^2 + k_3 x^3 \)
              • \( U(x) = A e^{-bx} \)
              • \( U(x) = {a constant} \)

            • 6.
              In a region, steady and uniform electric and magnetic fields are present. These two fields are parallel to each other. A charged particle is released from rest in this region. The path of the particle will be a

                • helix
                • straight line
                • ellipse
                • circle

              • 7.
                A flask contains a monoatomic and a diatomic gas in the ratio of \( 4:1 \) by mass at a temperature of \( 300K \). The ratio of average kinetic energy per molecule of the two gases is:

                  • \( 1:1 \)
                  • \( 2:1 \)
                  • \( 4:1 \)
                  • \( 1:4 \)

                • 8.
                  A sinusoidal voltage of amplitude \( 25 \) V and frequency \( 50 \) Hz is applied to a half-wave rectifier using a P-N junction diode. No filter is used, and the load resistor is \( 1000\Omega \). The forward resistance \( R_f \) of the ideal diode is \( 10\Omega \). The percentage rectifier efficiency is:

                    • \( 40\% \)
                    • \( 20\% \)
                    • \( 30\% \)
                    • \( 15\% \)

                  • 9.
                    A heater coil is cut into two equal parts and only one part is now used in the heater. The heat generated will now be

                      • four times
                      • doubled
                      • halved
                      • one fourth

                    • 10.

                      Consider an electric field \( \mathbf{E} = E_0 \hat{x} \), where \( E_0 \) is a constant. The flux through the shaded area (as shown in the figure) due to this field is: 
                       

                        • \( 2E_0 a^2 \)
                        • \( \sqrt{2} E_0 a^2 \)
                        • \( E_0 a^2 \)
                        • \( \frac{E_0 a^2}{\sqrt{2}} \)

                      • 11.

                        If a charge \( q \) is placed at the centre of a closed hemispherical non-conducting surface, the total flux passing through the flat surface would
                         

                          • zero
                          • \( \frac{q}{2\epsilon_0} \)
                          • \( \frac{q}{4\epsilon_0} \)
                          • \( \frac{q}{2\pi\epsilon_0} \)

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