Thank you FULL SCREEN PRINTER VERSION BACK NEXT Chapter 19, Problem 38 GO An ele

Question

Thank you FULL SCREEN PRINTER VERSION BACK NEXT Chapter 19, Problem 38 GO An electron is released from rest at the negative plate of a parallel plate capacitor and accelerates to the positive plate (see the drawing). The plates are separated by a distance of 1.3 cm, and the electric field within the capacitor has a magnitude of 2.4 x 106 v/m. What is the kinetic energy of the electron just as it reaches the positive plate? Electric field Electron KEpositive" the tolerance is +/-5% SHOW HINT GO TUTORIAL LINK TO TEXT LINK TO TEXT

Explanation / Answer

Solution :-

Charge of the electron ( q ) = 1.6 * 10-19C

    Magnitude of the electric field ( E ) = 2.4 *106 V/m

   Seperation distance ( d ) = 1.7 cm = 0.013m

    As we know that

Work done by electric field ( W ) = change in thekinetic energy ( K )

    F . d = K

    K = E q . d

         = (2. 4*106 V/m) (1.6 * 10-19 C) (0.013m)

         = 4.992 ×10-15 J

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