Electrostatics Quiz

An elektroskop is a device used to determine whether an object contains a charge or not. It works by detecting the presence of an electric field around the object. When the object is charged, the elektroskop will show a deflection or movement of its indicator. Therefore, an elektroskop is the correct tool to use for determining the presence of a charge in an object.

The given statement is describing the concept of "Fluks listrik" which refers to the electric flux. Electric flux is a measure of the number of electric field lines that pass through a given surface. It is a concept used in electromagnetism to understand the flow of electric fields and is calculated by taking the dot product of the electric field and the area vector of the surface.

The direction of the negative electrical charge field is inwardly radial. This means that the field lines originate from the negative charge and point towards it in a radial direction. This is because negative charges repel each other, so the field lines move away from other negative charges and towards positive charges.

The energy stored in a capacitor is given by the formula E = 1/2 CV^2, where E is the energy, C is the capacitance, and V is the voltage. In this case, the energy is given as 3.6 x 10^-5 Joule and the capacitance is 2 μF. Plugging these values into the formula, we can solve for V. Rearranging the formula, we get V = sqrt(2E/C). Substituting the given values, we find V = sqrt(2 * 3.6 x 10^-5 / 2 x 10^-6) = sqrt(36) = 6 volt. Therefore, the correct answer is 6 volt.

When a dielectric material is inserted into a capacitor, it increases the capacitance of the capacitor. The capacitance of a capacitor with a dielectric material can be calculated using the formula C = C0 * εr, where C is the capacitance with the dielectric, C0 is the capacitance without the dielectric, and εr is the relative permittivity or dielectric constant of the material. In this case, the initial capacitance without the dielectric is 10 μF. When the mica with a dielectric constant of εr = 5 is inserted, the capacitance becomes 10 μF * 5 = 50 μF. Therefore, the correct answer is 50 μF.

When plastic is rubbed against dry hair or wool, the plastic comb becomes negatively charged due to an excess of electrons transferred from the hair or wool. At the same time, the hair or wool becomes positively charged. This charge imbalance between the comb and the hair/wool creates an electrostatic attraction, causing the comb to attract small pieces of paper.

The capacitor with the largest capacitance is C4 because it has the largest area of the parallel plates among all the capacitors. The capacitance of a capacitor is directly proportional to the area of its plates. Therefore, C4 has the largest capacitance compared to C1, C2, C3, and C5.

Based on the given information, A attracts B and repels C. Since C attracts D and C is positively charged, it can be inferred that D is negatively charged. Therefore, A must be positively charged in order to attract B. This leaves us with the answer choice A positif, B negatif, and D negatif.

The point that has a zero electric field is located at a distance of 0.40 m from q1 in the direction towards the right.

The electric potential at a distance from a charged particle is given by the equation V = kQ/r, where V is the electric potential, k is the Coulomb's constant, Q is the charge of the particle, and r is the distance from the particle. In this case, the charge of the hydrogen atom is 1.6 x 10^-19 C and the distance is 2.4 x 10^-10 m. Plugging these values into the equation, we get V = (9 x 10^9 Nm^2/C^2)(1.6 x 10^-19 C)/(2.4 x 10^-10 m) = 6 volts. Therefore, the electric potential at a distance of 2.4 x 10^-10 m from the hydrogen atom is 6 volts.

Based on the given diagram, we can conclude that qA is positive and the magnitude of qA is equal to the magnitude of qB.

When the distance between the charges is doubled to 2r, the repulsive force between them decreases. This is because the repulsive force between two charges is inversely proportional to the square of the distance between them. Therefore, if the distance is doubled, the force will be reduced by a factor of (1/2)^2 = 1/4. Hence, the repulsive force becomes 1/4 F.

The correct answer is E. A cos 60. This is because the flux of an electric field through a surface is given by the formula Φ = E * A * cosθ, where E is the electric field strength, A is the area of the surface, and θ is the angle between the electric field and the surface. In this case, the electric field is homogeneous and is passing through the surface at an angle of 30 degrees. Since the cosine of 30 degrees is equal to the cosine of 60 degrees, the answer is E. A cos 60.

The correct answer is "Mengatur besarnya arus listrik" because capacitors do not have the ability to regulate the magnitude of electric current. Capacitors are primarily used to store and release electrical energy, as well as to filter out certain frequencies in electronic circuits. They are not designed to control or adjust the magnitude of electric current flowing through a circuit.

The charge in a capacitor is given by the equation Q = CV, where Q is the charge, C is the capacitance, and V is the voltage difference across the capacitor. In this case, the capacitance of C2 is given as 2 μF and the voltage difference across C2 is 12 volts. Plugging these values into the equation, we get Q = (2 μF)(12 V) = 24 μC. Therefore, the charge in C2 is 24 μC.