Electrostatics — Part 1 MCQs with Explanations

Option A: low permittivity A dielectric with low permittivity would result in a capacitor with lower capacitance. This is not desirable for most applications where higher capacitance is needed. Option B: high permittivity A dielectric with high permittivity increases the capacitance of a capacitor. Higher permittivity means the dielectric material can store more electric energy for the same electric field, making it a better choice for capacitors. Option C: permittivity same as that of air Using a dielectric with the same permittivity as air would not provide any advantage over using air itself. It would not enhance the capacitance of the capacitor significantly. Option D: permittivity slightly more than that of air While a dielectric with permittivity slightly more than that of air would increase the capacitance somewhat, it would not be as effective as a dielectric with high permittivity. Conclusion: To make a capacitor with higher capacitance, it is better to select a dielectric with high permittivity, making Option B the correct answer.

Option A: Mica Mica capacitors are known for their high breakdown voltage capability. They can withstand higher voltages compared to many other types of capacitors. Option B: Paper Paper capacitors typically have moderate breakdown voltage ratings. They are used in applications where moderate voltage levels are encountered. Option C: Ceramic Ceramic capacitors usually have breakdown voltages that vary widely depending on their construction. They can be designed for various voltage ratings, from low to high. Option D: Electrolytic Electrolytic capacitors generally have the lowest breakdown voltage among the options listed. They are designed for applications where moderate to low voltage levels are involved. Exceeding their breakdown voltage can cause permanent damage. Conclusion: Among the given options, electrolytic capacitors typically have the least value of breakdown voltage, making Option D the correct answer.

Here's the explanation for the capacitance problem: The question is about finding the capacitance of a capacitor. First, remember the basic formula that connects charge (Q), voltage (V), and capacitance (C): Q = C * V where: Q is the charge stored in the capacitor (in Coulombs). C is the capacitance (in Farads). V is the voltage across the capacitor (in Volts). We are given: Q = 2000 μC (which is 2000 * 10 -6 Coulombs) V = 200 V We need to find C. Rearrange the formula to solve for C: C = Q / V Substitute the given values: C = (2000 * 10 -6 ) / 200 C = 10 * 10 -6 F C = 10 μF (microFarads) So, the correct answer is Option B: 10 uF .

First, let's understand the key concepts: * Electric Field (E): This is the force exerted on a charge due to the voltage difference. * Force on an Electron (F): An electron experiences a force in an electric field. * Acceleration (a): This is the rate of change of velocity caused by the force. Here's how to solve the problem step-by-step: 1. Find the Electric Field (E): * The electric field between two parallel plates is uniform and calculated as E = V/d , * where V is the potential difference (100 V) and d is the distance between the plates (10 mm = 0.01 m). * So, E = 100 V / 0.01 m = 10000 V/m. 2. Find the Force on the Electron (F): * The force on an electron in an electric field is F = qE , * where q is the charge of an electron (approximately 1.602 x 10 -19 Coulombs). * So, F = (1.602 x 10 -19 C) * (10000 V/m) = 1.602 x 10 -15 N. 3. Find the Acceleration (a): * Using Newton's second law, F = ma , where m is the mass of the electron (approximately 9.109 x 10 -31 kg). * So, a = F/m = (1.602 x 10 -15 N) / (9.109 x 10 -31 kg) ≈ 1.76 x 10 15 m/s 2 . Therefore, the correct answer is: Option D: 1.76 x 10 15 m/s 2

When a parallel-plate capacitor is charged using a constant voltage source and then disconnected from the source, the charge Q on the capacitor remains constant because there is no conducting path for the charge to escape. The capacitance of a parallel-plate capacitor is given by: C = εA/d where d is the separation between the plates. When the plates are separated apart using insulated handles, the distance d increases . Therefore, the capacitance C decreases . Using the capacitor relation: Q = CV Since Q remains constant and C decreases , the voltage V must increase to maintain the same charge: V = Q/C Thus, increasing the plate separation causes the voltage across the plates to increase . Why the other options are incorrect: Option B — Voltage decreases: Incorrect. Since capacitance decreases while charge remains constant, voltage increases. Option C — Charge decreases: Incorrect. The plates are handled with insulated handles and the capacitor is disconnected from the source, so the charge remains essentially constant. Option D — Charge increases: Incorrect. There is no external source connected to supply additional charge. Key Concept: For an isolated charged capacitor , increasing plate separation causes capacitance to decrease and voltage to increase , while the charge remains constant . Final Answer: Option A — Voltage across the plates increases.

The permittivity of Free Space ϵ 0 = 8.85✕ 10 -12 Farad/meter

This question asks about the dielectric constant , which is a material's ability to store electrical energy in an electric field. A lower dielectric constant means the material stores less energy. Let's look at the options: * Ceramics generally have a relatively high dielectric constant. * Oil has a lower dielectric constant compared to ceramics and glass. * Glass also has a relatively high dielectric constant. * Paper has a dielectric constant higher than oil. Therefore, oil has the least value of dielectric constant compared to the other materials listed. So the answer is Option B: Oil

Here's an explanation of the problem and solution: The question deals with the electric field intensity due to a point charge . The electric field intensity (E) due to a point charge (Q) at a distance (r) is given by the formula: E = kQ/r 2 , where k is a constant. This means the electric field intensity is inversely proportional to the square of the distance . So, if the distance is halved, the electric field intensity will be quadrupled (multiplied by 4). Initially, at 4 meters, the electric field intensity is 200 N/C. When the distance is reduced to 2 meters (which is half of 4 meters), the new electric field intensity will be 200 N/C * 4 = 800 N/C. Therefore, the correct answer is Option C: 800 N/C

Let's break down this problem step-by-step to understand how to find the electric field strength. What is Electric Field Strength? The electric field strength (often denoted by 'E') tells us how strong the electric force is at a particular point. Think of it like this: the higher the electric field strength, the stronger the force a charge would feel if placed there. The Relationship Between Potential and Electric Field When we have a uniform electric field (meaning the field is the same strength and direction everywhere), there's a simple relationship between the potential difference (voltage, 'V') and the electric field strength ('E'): E = V / d Where: * 'E' is the electric field strength (what we want to find) * 'V' is the potential difference (voltage) between the plates * 'd' is the distance between the plates Applying the Formula to Our Problem We're given: * V = 400 V (potential difference) * d = 4 mm = 0.004 m (distance between plates - we need to convert mm to meters!) Now, plug these values into the formula: E = 400 V / 0.004 m E = 100000 V/m E = 100 kV/m (Kilovolts per meter) Therefore, the correct answer is Option A: 100 kV/m

A dielectric is an insulating material that can withstand electric stress without breakdown. The superiority of a dielectric is judged mainly by its dielectric strength and insulating capability . Among the given options, glass has a very high dielectric strength and excellent insulating properties compared to air, paper, and bakelite. Air has the lowest dielectric strength, while paper and bakelite are moderate dielectrics but inferior to glass. Therefore, glass is the most superior dielectric among the given options, making Option B the correct answer.

zero is the correct answer to “The power dissipated in a pure capacitor is”. This question is standard in competitive examination syllabi for Electrostatics.

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