Find the maximum potential difference between two parallel conducting plates separated by 0.5 cm of air, given the maximum sustainable electric field strength in air to be 3.0 × 106 V/m. Hint The maximum potential difference between them is V.
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- (a) Determine the electric field strength between two parallel conducting plates to see if it will exceed the breakdown strength for air (3 ✕ 106 V/m). The plates are separated by 3.48 mm and a potential difference of 5515 V is applied. V/m(b) How close together can the plates be with this applied voltage without exceeding the breakdown strength? mmAn electrostatic paint sprayer has a 0.25 m diameter metal sphere at a potential of 25.0 kV with respect to the ground that repels paint droplets onto a grounded object. Part (a) What charge is on the sphere, in coulombs? Q = Part (b) What charge, in coulombs, must a 0.12 mg drop of paint have in order to arrive at the object with a speed of 10.0 m/s, starting from rest? q =An electric field pointing in the positive x direction has a magnitude that increases linearly with x as described by the functionEx = b x, where b = 5 N/C/m.Find the potential difference between thepoints on the x-axis at x = 1 m and x = 2 m.Answer in units of V.
- Try to help me with this question. no handwritten.A water molecule is built as an isosceles triangle; the two hydrogen atoms are separated by 1.5x10-10m, and the oxygen is a distance 1x10-10m from either of the hydrogens. Treating the hydrogens as point charges, each of +e, and the oxygen as a point charge of -2e, estimate the total electric potential energy stored in this molecule.The electric field strength between two parallel conducting plates separated by 3.40 cm is 7.10 ✕ 104 V/m. (a) What is the potential difference between the plates (in kV)? kV (b) The plate with the lowest potential is taken to be at zero volts. What is the potential (in V) 1.90 cm from that plate (and 1.50 cm from the other)? V