Free Electrostatics MCQs with Answers
831 Electrostatics MCQs from Physics, each with the correct answer and a written explanation of why it is correct. Free and unlimited, with no account needed.
Electrostatics describes forces between charges through Coulomb's law, electric fields and field intensity, then uses electric potential to measure energy per unit charge. Capacitors store charge and electrical energy, with capacitance depending on geometry and dielectric material, not simply on the amount of charge present.
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Read the Electrostatics notesFree MDCAT chapter notes with key terms831 questions · page 41 of 42
- A. Be fully charged in 1 second by a current of 1 Ampere
- B. Store 1 coulomb of charge at potential difference of 1 volt
- C. Gain 1 joule of energy when 1 coulomb of charge is stored on it
- D. Discharge in 1 second when connected across a resistor of resistance 3 ohms
Explanation: A capacitance of 1F produces 1V of potential difference for an electric charge of one coulomb (1C).
Correct answer: Store 1 coulomb of charge at potential difference of 1 volt- A. E√10
- B. E/√10
- C. 10E
- D. E/10
Explanation: The relationship between the electric field in the absence of the dielectric (E0) and the electric field with the dielectric (E) is given…
Correct answer: E/10- A. 4x
- B. 2x
- C. x/2
- D. x/4
Explanation: C = A𝛆 / d where C is capacitance, 𝛆 is the permittivity of free space, A is the area of plates, and d is the distance between them.
Correct answer: x/2- A. 400 J
- B. 4 J
- C. 0.2 J
- D. 200 J
Explanation: The energy stored in a capacitor can be calculated using the formula U = 0.5 * Q * V, where U is the energy, Q is the charge, and V is the…
Correct answer: 200 J- A. 2 F
- B. 4 F
- C. 0.5 F
- D. 0.25 F
Explanation: The potential difference across the parallel plate capacitor is 10V- (-10)=20VCapacitance= Q/V = 40/20 = 2 FaradAs it is numerical, it can…
Correct answer: 2 F806. When dielectric is removed from plates of a charged capacitor then electric field between plates:
- A. Remains unchanged
- B. Decreases
- C. Increases
- D. All of the above
Explanation: When the dielectric is removed from the plates of a charged capacitor, the electric field between the plates generally increases.
Correct answer: Increases- A. C - E
- B. E.C
- C. C + E
- D. C/E
Explanation: The capacitance of a parallel plate capacitor with a dielectric is given by the formula C = E * (ε₀ * A / d), where E is the dielectric…
Correct answer: C/E- A. 6
- B. 9
- C. 12
- D. 18
Explanation: When capacitors are connected in parallel, their equivalent capacitance is the sum of their individual capacitances.
Correct answer: 6- A. F-1=1/4πe0 x q1q2/r2 x r
- B. F = 1/4πe0 x q1q2/r2 x r
- C. F-1 = 1/4πe x q1q2/r2 x r
- D. F = 1/4πe x q1q2/r2 x r
Explanation: According to the expression of coulomb's law, the magnitude of Coulomb's force is directly proportional to the product of two point…
Correct answer: F = 1/4πe0 x q1q2/r2 x r- A. Lost electrons
- B. Lost protons
- C. Gained protons
- D. Gained electrons
Explanation: When a neutral body becomes positively charged, it means it has lost electrons.
Correct answer: Lost electrons811. The Ratio of electro force between two protons to that between two electrons is of the order of:
- A. 1042
- B. 1039
- C. 1036
- D. 1
Explanation: d) 1This option suggests that the ratio of the electrostatic force between two protons to that between two electrons is 1.
Correct answer: 1812. When 1012 electrons are received from a neutral metal sphere. The charge on the sphere becomes:
- A. 0.16uc
- B. -0.16uc
- C. 0.32uc
- D. -0.32uc
Explanation: The formula used is q (charge) = n x e so q = 1012 x 1.6x10-19 =1.6x10-7=0.16 ucAs it is numerical, it can have only one possible answer.
Correct answer: 0.16uc- A. Volt/meter
- B. Newton/coulomb
- C. Joule/coulomb_meter
- D. All of above
Explanation: Therefore, the correct answer is "All of the above" since all the options are valid units for electric intensity, each expressing a…
Correct answer: All of above- A. 1.6x10^-4
- B. 1.6x10^-8
- C. 1.6x10^-10
- D. 1.6x10^-11
Explanation: The force on an electron in an electric field can be calculated by using the formula:F = qEWhere 'q' is the charge of the electron.
Correct answer: 1.6x10^-11- A. 240 NC-1
- B. 24 NC-1
- C. 2.4 NC-1
- D. 2400 NC-1
Explanation: As we know, Potential gradient is given by E= -dV/dR So, 12/0.005 = 2400 V/m= 2400 N/C
Correct answer: 2400 NC-1- A. Potential energy
- B. Electric field intensity
- C. Electric potential difference
- D. Electron-volt
Explanation: The correct answer is Electric field intensity. The electric field intensity is defined as the negative gradient of electric potential…
Correct answer: Electric field intensity- A. Separation of the plates
- B. Separation and area of the plates
- C. Permittivity of the medium; separation of the plates
- D. Permittivity of the medium; separation and area of the plates
Explanation: The correct answer is Option A: Separation of the plates. The electric field strength between parallel conducting plates can be calculated…
Correct answer: Separation of the plates- A. N/C
- B. V/m
- C. J/C.m
- D. All of the above
Explanation: The units of the electric field in the SI system are newtons per coulomb (N/C), or volts per meter (V/m); in terms of the SI base units…
Correct answer: All of the above- A. Volt/metre
- B. Newton/coulomb
- C. Joule/Coulomb.metre
- D. volt.metre
Explanation: The electric field intensity is defined as the force experienced by a positive test charge divided by the magnitude of the charge.
Correct answer: Newton/coulomb- A. Electric field
- B. Electric flux
- C. Electric intensity
- D. Gravitational field
Explanation: SHOULD be B)A: An electric field is defined as any region around a charge in which an electric test charge would experience an electric…
Correct answer: Electric flux