Calculate the electrical potential energy in Joules, of two particles (3.561x10^0) nC and (7.22x10^0) nC, which are a distance (5.0287x10^0) m apart.
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- In a particular region, the electric potential is given by V O a. E = (y²z - 4y)i + xy²j + (2xyz – 4x)k O b. E = (-y²z + 4y)i xy²j + (−2xyz + 4x)k O c. E=(-y²z + 4y)i + (−2xyz + 4x) ĵ – xy²k O d. E= (y²z - 4y)i + (2xyz — 4x)j + xy²k -xy²z + 4xy. What is the electric field in this region?A particle with a charge of q = 12.0 µC travels from the origin to the point (x, y) = (20.0 cm, 50.0 cm) in the presence of a uniform electric field E = 2901 V/m. Determine the following. (a) the change in the electric potential energy (in J) of the particle-field system J (b) the electric potential difference (in V) through which the particle movesA -7.8 C charge is moving in a electric potential given by V(x) = 6 x4 (V). The particle begins at x = 9 m and ends at x = 19 m. Calculate the change in electrical potential energy of the charge, in J. (Please answer to the fourth decimal place - i.e 14.3225)
- Two charges, qi =-14.0 mC and q2 =8.4 mC are separated by distance r =6.6 m. What is the change in the electric potential energy of the system of these two charges if the separation changes to 0.7r? Provide your answer in kilojoules, with a precision of one place after the decimal.A particle of mass 3 × 10−5kg and charge1 µC moves downward from point A to pointB a distance of 3 m in the earth’s gravitational field. The kinetic energy of the particledecreases by 3.6 mJ during this movement.What is the potential difference VB − VA?The acceleration of gravity is 9.8 m/s2.Answer in units of kV.An alpha particle, which has charge 3.204 ✕ 10−19 C, is moved from point A, where the electric potential is 2.30 ✕ 103 J/C, to point B, where the electric potential is 3.50 ✕ 103 J/C. Calculate the work in electron volts done by the electric field on the alpha particle.
- An electron moving parallel to the x axis has an initial speed of 3.70 3 106 m/s at the origin. Its speed is reduced to 1.40 3 105 m/s at the point x 5 2.00 cm. (a) Calculate the electric potential difference between the origin and that point. (b) Which point is at the higher potential?Please asapfind magnitude of EF at (3.0x +5.1y - 2.5z) m. electric potential is (x.y.z)=5.5•x•(y^2)•z