The electric field intensity between two charged plates is 1.5*10^3 N/C. The plates are 0.060 meters apart...

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The electric field intensity between two charged plates is 1.5*10^3 N/C. The plates are 0.060 meters apart...

The electric field intensity between two charged plates is 1.5*10^3 N/C. The plates are 0.060 meters apart. What is the electric potential difference, in volts, between the plates?

Please help me with this problem.
I don't understand it.

2007-04-24 12:49:07 · 4 answers · asked by vicky p 1 in Science & Mathematics ➔ Physics

4 answers

The units N/C are equivalent to volts/meter. Therefore this times distance in meters gives volts. 0.06*1.5*10^3 = 90.0 volts

2007-04-24 12:55:30 · answer #1 · answered by gp4rts 7 · 0⤊ 0⤋

Charged Plates

2016-12-18 07:43:03 · answer #2 · answered by Anonymous · 0⤊ 0⤋

V = Ed

V = 1.5 x 10^3 * .06

V = 90V

2007-04-24 12:52:37 · answer #3 · answered by 7 · 0⤊ 0⤋

8e-4N to ward the positive plate

2016-05-17 23:47:46 · answer #4 · answered by linnie 3 · 0⤊ 0⤋