The electric field strength between the plates of a simple air capacitor is equal to the voltage across the plates divided by the distance between them. When a voltage of 108.V is put across the plates of such a capacitor an electric field strength of 3.0kVcm Write an equation that will let you calculate the distance d between the plates. is measured.Your equation should contain only symbols. Be sure you define each symbol.

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Current Attempt in Progress Your answer is partially correct. A 6.44-kg box is sliding across the horizontal floor of an elevator. The coefficient of kinetic friction between the box and the floor is 0.271. Determine the kinetic frictional force that acts on the box when the elevator is (a) stationary, (b) accelerating upward with an acceleration whose magnitude is 1.37 m/s², and (c) accelerating downward with an acceleration whose magnitude is 1.37 m/s². (a) Number i (b) Number i (c) Number eTextbook and Media Hint Save for Later Units N Units N Units N Attempts: unlimited Submit Answer

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Table 4. V U K= 3.5 (paper) (RC) (pF) (V) (J) Connected Battery 0.34*10^-12C 0.22*10^-12F 1.5 v 0.25*10^-12J without dielectric Disconnected 0.34*10^-12 C 0.220*10^-12F 1.5 V 0.25*10^-12J Battery with dielectric

You are working on an electronics project that requires a variety of capacitors, but you have only five capacitors available, each with a capacitance of 170 nF. Part A What is the maximum capacitance that you can produce with these five capacitors? Express your answer in nanofarads. Cmax = Submit Part B Cmin = VG ΑΣΦ Submit Request Answer What is the minimum capacitance, other than zero, that you can produce? Express your answer in nanofarads. VG ΑΣΦ ? Request Answer nF nF

In the circuit below, R1 = 300, R2 = 150, R3 = 525, R1 = 300, V = 13. Find the voltage across R1. Please give %3D your answer in Volts. R. R. R,

Could you help me to solve the following problem step by step?Thank you very much in advance.6. A) How is capacitance defined? B) To a capacitor with air between its plates with an area of 3 cm2 and a separation between them of 1 mm. If a potential difference of 5 V is applied, calculate a) the electric field generated, b) the surface charge density c) the capacitance d) A dielectric of dielectric constant =3 d) What is the capacitance now?

Coulomb's Law 1. ke is the Coulomb force constant. It is the proportionality constant in the Coulomb Force Equation which describes electrostatic forces. Fe = k9192 r2 a. The magnitude of ke is 8.99 x 10%. What are its units? Show your work. b. Using the equation above describe how the force between the two charges changes as the distance increases.