Question

Q 5 : Two parallel plates are charged. The separation between the plates is 2 cm and the magnitude of the electric field between the plates is 500 Volt/meter. What is the change in potential energy of the proton when accelerated from the positively charged plate to the negatively charged plate. Q 6: Two point charges are separated by a distance of 10 cm. Charge on point A =+9 ?C and charge on point B = -4 ?C. k = 9 x 10^9 Nm^2C^-2, 1 ?C = 10^-6 C. What is the change in electric potential energy of charge on point B if accelerated to point A ?

          Q 5 : Two parallel plates are charged. The separation between the plates is 2 cm and the magnitude of the electric field between the plates is 500 Volt/meter. What is the change in potential energy of the proton when accelerated from the positively charged plate to the negatively charged plate.
Q 6: Two point charges are separated by a distance of 10 cm. Charge on point A =+9 ?C and charge on point B = -4 ?C. k = 9 x 10^9 Nm^2C^-2, 1 ?C = 10^-6 C. What is the change in electric potential energy of charge on point B if accelerated to point A ?

Q 5 : Two parallel plates are charged. The separation between the plates is 2 cm and the magnitude of the electric field between the plates is 500 Volt/meter. What is the change in potential energy of the proton when accelerated from the positively charged plate to the negatively charged plate.
Q 6: Two point charges are separated by a distance of 10 cm. Charge on point A =+9 ?C and charge on point B = -4 ?C. k = 9 x 10^9 Nm^2C^-2, 1 ?C = 10^-6 C. What is the change in electric potential energy of charge on point B if accelerated to point A ?

Added by Lauren P.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Vishal Gupta

11/14/2021

Step 1

The change in electric potential is given by ΔV = E*d, where E is the electric field and d is the distance. For a proton, the charge q = 1.6 * 10^-19 C. So, ΔPE = q*E*d = (1.6 * 10^-19 C) * (500 V/m) * (2 * 10^-2 m) = 1.6 * 10^-21 Joules. Show more…

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Q 5 : Two parallel plates are charged. The separation between the plates is 2 cm and the magnitude of the electric field between the plates is 500 Volt/meter. What is the change in potential energy of the proton when accelerated from the positively charged plate to the negatively charged plate. Q 6: Two point charges are separated by a distance of 10 cm. Charge on point A =+9 ΑC and charge on point B = -4 ΑC. k = 9 x 10^9 Nm^2C^-2, 1 ΑC = 10^-6 C. What is the change in electric potential energy of charge on point B if accelerated to point A ?