The electric potential V is given as a function of distance ×( metre ) by V=(5x2+10x−4) volt.Question: Value of electric field at x=1 m

If the electric potential along the x-axis is V = 500 x2𝑥2 V, where x is in meters, what is Ex𝐸𝑥 at x = 0.500 m? 250. V/m -250. V/m -500. V/m 500. V/m

The electric field at a point due to a point charge is 30 N/C and the electric potential at that point is 15 J/C respectively. What is the distance of the point from the charge and the magnitude of the charge?0.1 m, 0.43 nC0.4 m, 0.77 nC0.5 m, 0.83 nC0.6 m, 0.31 nC

Suppose an electric potential is described by V(x, y, z) = −(5x2 + y + z)𝑉(𝑥, 𝑦, 𝑧) = −(5𝑥2 + 𝑦 + 𝑧) in volts. Which of the following expressions describes the associated electric field, in units of volts per meter?(a)  E⃗ =5x^+2y^+2z^𝐸→=5𝑥^+2𝑦^+2𝑧^ (b)  E⃗ =10xx^𝐸→=10𝑥𝑥^ (c)  E⃗ =5xx^+2y^𝐸→=5𝑥𝑥^+2𝑦^ (d)  E⃗ =10xx^+y^+z^𝐸→=10𝑥𝑥^+𝑦^+𝑧^ (e)  E⃗ =0𝐸→=0 The correct response is

A 0.3 uC of source charge is placed in the electric field. Calculate it’s electric potential at a distance 5 cm away from a test charge.Group of answer choices54x10^-3 J/C54000 J10.8x10^7 J54x10^3 J/C

Determine the electric potential produced by a point charge of 6.80x10-7 C at a distance of 2.60 cm.