The voltage in a region of space is decried by the following equation. V=50-2x^2

Question

The voltage in a region of space is decried by the following equation. V=50-2x^2+6y^(2/3) where x and y are in meters. use this equation to ans the following: a) What is the electric field in unit vector form, at location (4.0,1.0) meters? b) This voltage is made by some charge distribution. Does that distribution contain only positive charge, only negative charge or both types of charge? fully explain your ans. c) If an electron with a 9.11*10^-31 kg mass is released from rest at (4.0,1.0) meters, what will its speed once it reaches the location, (2.0,2.0) meters?

Explanation / Answer

a) E = -d(V)/d(x) i^ + -d(V)/d(y) j ^ = 4x i^ - 4/3 {y^(-1/3) } j^ at (4, 1) E = 16 i - 4/3 j b) The distribution contains some negative charge also because at some regions , the voltage is negative which can only occur when negative charge is present. c) Change in potential = 27.52V change in potential energy = qV= 1.6* 10^-19* 27.52 = change in kinetic energy = 0.5* 9.11* 10^-31 * VELOCITY^2 => velocity = 3.1* 10^6 m/sec

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