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I need help with Physics please!!?

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I need help with Physics please!!?

───── ( 15 ) ─────

| F | = | E | x | q |

Charge of electron q = 1.608 x 10^(-19) C

| F | = 5 x 10^3 x 1.608 x 10^(-19) = 8.04 x 10^(-16) N

which is approximately equal to (B) 8.0 x 10^(-16) N

───── ( 21 ) ─────

Potential difference between two points is the work done in moving unit positive charge from one point to the other.

Work done in transferring 5 C of charge = 20 J

Work done in transferring 5 C of charge = 4.0 J

That is the potential difference between the two points is equal to 4.0 V ───── (D)

───── ( 24 ) ─────

This question is similar to the above question. At infinity, the potential is assumed to be zero.

Potential energy of the 2C charge at that point W = 8.0 J

Electric potential at that point = W/q = 8J / 2C = 4.0 V ── (D)

───── ( 25 ) ─────

This is similar to ( 15 )

Elementary charge is nothing but the magnitude of charge on the electron

| F | = 3 x 10^3 x 1.6 x 10^(-19) = 4.8 x 10^(-16) N ─── (D)

Electric Potential

Electric Potential – A constant electric field is the field in between two parallel sheets equally but oppositely charged. If the positive sheet is on the left and the negative on the right, as shown, the direction of E is to the right or along the (+x) axis.Since E is constant, so is F, the force on any point charge q placed in E.The Work done on +q by constant F as it is pushed to the right varies linearly withDetermine a) the electric field strength at a point X at a distance 20 cm from a point charge Q = + 6µC. (1.4 x 10 6 N/C) b) the electric force that acts on a point charge q = -0.20 µC placed at point X. (0.28 N towards Q). + + – – 2 q 1 q 20 cm X Two point charges, q 1 = +2.0 C and q 2 = -3.0 C, are separeted by a distance of 40 cm, as shown in figure above.I need help with Electrostatics 15) What is the magnitude of the electric force acting on an electron located in an electric field with an intensity of 5 x 10^3 Newtons per coulomb A) 3.2 x 10^-23N B) 8.0 x 10^-16N C) 5.0 x 10^3N D) 3.2 x 10^22N 21) If 20 Joules of work is done in transferring 5.0 Coulombs of charge between two points, the potential difference between these two points is A

Calculate the electric field strength at point A What – Figure 2. The magnitude of the electrostatic force F between point charges q 1 and q 2 separated by a distance r is given by Coulomb's law. Note that Newton's third law (every force exerted creates an equal and opposite force) applies as usual—the force on q 1 is equal in magnitude and opposite in direction to the force it exerts on q 2(A) It is a maximum at point B. (B) It is the same at points A, B, and C. (C) It is a maximum at point C. (D) It is zero at point B. An electron is placed between two oppositely charged parallel plates as shown in the diagram below. — Electron 20. 21. 22. Two parallel aluminum plates are connected to a source of potential as shown in the diagram.z Electric Field due to a Point Charge To calculate the electric field at any point at a distance r in space from a point charge q, imagine a test charge q 0 placed at that point. The magnitude of the electric force on q 0 is: Thus the magnitude of the electric field due to the point charge is: Like electric force, electric field is also vector quantity and has the same direction as the

Calculate the electric field strength at point A What

I need help with Physics please!!? | Yahoo Answers – A) Find the electric field (magnitude and direction) at each of the following points on the x axis. i) x-2.00m; ii) x=1.2m iii) x= -0.2m. b) Find the net electric force that the two charges would exert on an electron placed at each point in part a). Answer: A) ⃗ = ⃗ 1+ ⃗ 2 𝑖= 𝑘 1 1𝑖 2− 𝑘 2 2𝑖 2 =9∗109∗(2∗ 10−9What is the magnitude of an electric field in which the electric force on a proton is equal in magnitude to its weight? Use 1.67×10−27 kg for the mass of a proton, 1.60×10−19 C for the magnitude of the charge on an electron, and 9.81 m/s2 for the magnitude of the acceleration due to gravity.At the mid-point they cancel and the electric field at q is 0. The potential at point q caused by Q1 is just a number. The potential from Q2 will be the same number (note the r on the bottom of the equation is a magnitude, it doesn't change sign). So the potentials add rather than cancel.

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