**Question:**A potential difference of 2-V is applied between two large parallel metal plates separated by a distance of 2.0 mm. The strength of the electric field between the plates is:- ○ 1000 V/m- ○ 6000 V/m- ○ 20 V/m- ○ zero**Explanation:**This question concerns the calculation of the electric field between two parallel plates when a voltage is applied. The electric field (\(E\)) in such a scenario is given by the formula:\[ E = \frac{V}{d} \]where \(V\) is the potential difference (voltage) and \(d\) is the separation distance between the plates.Given:- \(V = 2\) V- \(d = 2.0\) mm = 0.002 mPlug the values into the formula:\[ E = \frac{2}{0.002} = 1000 \text{ V/m}\]Thus, the strength of the electric field between the plates is 1000 V/m.

Write the given values with the suitable variables. V=2 Vd=2 mm=2×10-3 m Here, V is the potential…

A potential difference of 2-V is applied between two large parallel metal plates separated by a distance of 2.0 mm The strength of the electric field between the plates is: O 1000 V/m 6000 V/m 20 V/m zero

A potential difference of 2-V is applied between two large parallel metal plates separated by a distance of 2.0 mm The strength of the electric field between the plates is: O 1000 V/m 6000 V/m 20 V/m zero

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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