The electric potential V in the space between twoflat parallel plates 1 and 2 is given (in volts) by V =1260x2, wherex (in meters) is the perpendicular distance from plate 1.What is the magnitude and direction of the electric field atx = 3.9 cm? (Take thedirection perpendicular to and away from plate 1 to bepositive.) The electric potential V in the space between twoflat parallel plates 1 and 2 is given (in volts) by V =1260x2, wherex (in meters) is the perpendicular distance from plate 1.What is the magnitude and direction of the electric field atx = 3.9 cm? (Take thedirection perpendicular to and away from plate 1 to bepositive.)
Explanation / Answer
The electric potential is given by V = 1260x2 The electric field is calculated at a distance of x = 3.9 cm =3.9 * 10-2 m Therefore,the magnitude of the electric field is E = V * x or E = 1260x2 * x = 1260 * x3 or E = 1260 * (3.9 * 10-2)3 or E = 0.075 N/C The direction of the electric field is from the positive plateto the negative plate.This is because the electric field isdirected from the positive charge to the negative charge. or E = 0.075 N/C The direction of the electric field is from the positive plateto the negative plate.This is because the electric field isdirected from the positive charge to the negative charge.