Consider the action S=FF"d²x. Vary the potential according to A → A+ do where is a scalar field. Determine the variation in the action.
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- In two different situations a proton is released between the plates of a charged capacitor (uniform E-field). In the first situation, it is released with some initial velocity and travels from position 1 to position 2. In the second situation, it is released with some initial velocity and travels from position 1 to position 3. In both situations the initial speed is the same. 1. The potential energy change () : A. is equal for both paths B. is greater from 1 to 2 C. is greater from 1 to 3 D. cannot be determined by given information 2. The kinetic energy of the proton () : A. increases from 1 to 3, but decreases from 1 to 2 B. increases from 1 to 2, but decreases from 1 to 3 C. increases form 1 to 2 and 1 to 3 D. does not change along these paths 3. At the end of each trajectory (): A. the speed of first proton is greater B. the speed of second proton is greater C. the protons have same speed D. cannot be determined speed by given informationA parallel plate capacitor has a surface charge density of σ = 0.0046 C/m2 on one plate. A small sphere charged to Q = 5.2E-06 C is moved a distance d = 0.17 m between the plates. Refer to the figure, which is not drawn to scale. At what speed, in meters per second, is the sphere moving when it reaches the distance d, if its mass is 0.10 kg and it starts from rest?A particle of mass 6.4 g and charge 21.5 µC is moving in an electric potential field V(x, y) = C₁ x - C₂ ⋅ y² + c3・y.x², C1 • where c₁= 65 V/m, c2= 45 V/m², and c3= 20 V/m³. Find the electric field acting on the particle as a function of its position. Use V/m and meters for the units, but do not put them explicitly in Ē(x, y). The x-compoment of the E-field, Ex(x,y) = Units Select an answer ✓ The y-compoment of the E-field, Ey(x, y) = What is the magnitide of the particle's acceleration at x = 1 m and y = -1 m? The acceleration, a = Units Select an answer ✓ Units Select an answer ✓
- The electrostatic potential in all space is given here as a function of x, y and z. Find the electric field Ē(x.y,z) function and express your answer using unit vectors. V(x,y,z)= 4πεο 4 TE, Jx + y² +(z-d)*Use the worked example above to help you solve this problem. A proton is released from rest at x = -2.60 cm in a constant electric field with magnitude 1.51 x 10³ N/C, pointing in the positive x-direction. (a) Calculate the change in potential energy when the proton reaches x = 5.35 cm. (b) An electron is now fired in the same direction from the same position. What is its change in electric potential energy if it reaches x = 12.60 cm? J (c) If the direction of the electric field is reversed and an electron is released from rest at X = 4.00 cm, by how much has its electric potential energy changed when it reaches x = 7.50 cm? EXERCISE Find the change in electric potential energy associated with the electron in part (b) as it goes on from x = 0.126 to x = -0.018 m. (Note that the electron must turn around and go back at some point. The location of the turning point is unimportant, because changes in potential energy depend only on the end points of the path.) ΔΡΕ = JFind the energy stored in a uniformly charged solid sphere of radius R and charge q. Show that you can do it three different ways: (a) Use the equation W = 1/2 ∫ρV dτ. (b) Use the equation W = ε0/2 ∫ E2 dτ. Don’t forget to integrate over all space.(c) Use the equation W = ε0/2 ( ∫ E2 dτ + ∮ VE · da). Take a spherical volume of radius a. What happens as a → ∞?
- The drawling shows a plot of the electric potential V versus the displacement s. Determine the electric field in the region from 0.2 to 0.6 m. Be sure to include the proper + or - sign.How far below the axis has the electron moved when it reaches the end of the pla Express your answer in meters. Cathode-ray tubes (CRTS) are often found in oscilloscopes and computer monitors. In (Figure 1) an electron with an initial speed of 6.70x10° m/s is projected along the axis midway between the deflection plates of a cathode-ray tube. The potential difference between the two plates is 24.0 V and the lower 8.45x10-3 m plate is the one at higher potential. Submit Previous Answers v Correct Part F At what angle with the axis is it moving as it leaves the plates? Express your answer in degrees. 0 = 15.7 0 Submit Previous Answers v Correct Part G Figure < 1 of 1 How far below the axis will it strike the fluorescent screen S? Express your answer in meters. 2.0 cm S D = m 不6.0 cm米 12.0 cm Submit Request Answer15V IDV In the attached pic, equipotential lines with their electric potential values are given 1. Draw the electric field vector at each of the points A, B & C 2. Calculate the change in Electrical Potential Energy when a proton (with positive charge +e) moved from