Question

The electric field strength between two parallel conducting plates separated by 4.20 cm is 6.80 x 104 v/m. (a) What is the potential difference between the plates (in kV)2 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 2.30 cm from the other)? Additional Materials Reading .-2 points oscoPhys2016 19.2.WA016 0/20 Submissions Used A proton is acted on by an uniform electric fleld of magnitude 443 N/C pointing in the positive y direction. The particle is initially at rest (a) In what direction will the charge move? Select (b) Determine the work done by the electric field when the particle has moved through a distance of 2.35 cm from its initial position. (c) Determine the change in electric potential energy of the charged particle (d) Determine the speed of the charged particle. m/s

Please complete both

Answer 1

a)

V = E d = 6.8* 10^4* 0.042

V = 2856 V

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b)

V = 6.8* 10^4* 0.019

V = 1292 V

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plz post other problem separately

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