Most Repeated MDCAT Physics: Electrostatics MCQs

Explanation

The energy stored in a capacitor is given by the equation: W = Q2 / 2C When the charge Q is increased to 2Q, the new energy W' is: W' = (2Q)2 / 2C = 4Q2 / 2C = 4W Thus, the stored energy becomes four times the original energy, making Option D (4W) the correct choice. The other options misrepresent the relationship between charge and stored energy, as they do not account for the square relationship.

Explanation

The correct answer is that the energy supplied by the cell to the charge carriers is derived from the conversion of chemical energy into electrical energy. This process occurs in both primary and secondary cells, where chemical reactions within the cell generate electrical energy that can be used to power devices. Option A is incorrect because endothermic reactions and the conversion of heat energy into chemical energy are not the processes occurring in conventional electrical cells. Option C is incorrect as it describes the function of solar cells, which are distinct from the electrochemical cells typically used in circuits.

Explanation

The capacitance (C) of a capacitor depends on three main factors:1. *Area of the plates (A)*: Increasing the area of the plates increases the capacitance, as more charge can be stored.2. *Distance between the plates (d)*: Decreasing the distance between the plates increases the capacitance, as the electric field between them becomes stronger.3.

Explanation

It is the amount of electricity that a 1-ampere (A) current carries in one second (s). A quantity of 1 C is equal to the electrical charge of approximately 6.24 x 1018 electrons or protons. He, therefore, concluded that this minimum value of the charge is the charge on an electron. A given droplet between the two plates could be suspended in air if the gravitational force F = mg acting on the drop is equal to the electrical force F, =qE. The F can be adjusted equal to F, by adjusting the voltage. In this case, we can write, Fe=Fg or qE=mg If V is the value of p.

Explanation

Recall that the effective capacitance of capacitors in series is the reciprocal of the sum of reciprocals of individual capacitors (just like resistors in parallel)Therefore,1/Ceffective = 1/C1 + 1/C21/Ceffective = ½ +¼ 1/Ceffective = ¾C effective= 4/3 µF Hence option D is the correct answer.

Explanation

The change in electric potential with respect to distance is ΔV/Δr which represent s potential gradient. Hence, option A is correct.

Explanation

The amount of energy required in moving an electron of charge by application of 1V potential difference equals to 1 eV. STB Pg-64 Hence, option D is correct.

Explanation

Coulomb's law represents the relation between amount of charges and force between them. STB Pg-40 Hence, option C is correct.

Explanation

The correct answer is Option A: Separation of the plates. The electric field strength between parallel conducting plates can be calculated using the formula E = V/d, where V is the potential difference and d is the distance between the plates. Therefore, knowing the separation of the plates is the additional information needed to determine the electric field strength in this scenario.Options B, C, and D include unnecessary information such as the area of the plates and permittivity of the medium, which are not directly required to find the electric field strength between the plates.

Explanation

A capacitance of 1F produces 1V of potential difference for an electric charge of one coulomb (1C). This statement also fully agrees with the formula C=Q/V.

Explanation

The correct answer is that the potential must decrease (Option C). Electric field lines are oriented from higher to lower electric potential, indicating that potential decreases along the direction of the field. Options A and B are incorrect because electric field lines do not provide information about the magnitude of the electric field; thus, strength may vary in any direction. Option D is incorrect because Option C accurately describes the relationship between electric field lines and potential.

Explanation

The total capacitance of capacitors in series is found using the formula: 1/Ctotal = 1/C1 + 1/C2. For C1 = 2μF and C2 = 4μF, we get:1/Ctotal = 1/2 + 1/4 = 2/4 + 1/4 = 3/4So, Ctotal = 4/3 μF.This shows that the correct answer is 4/3 μF, making Option C correct. The other options result from incorrect calculations or misunderstandings of the series capacitor formula.