Four point charges, each with a magnitude of 1.00 C, are placed at the corners of a 45 rhombus with sides of length 1.0 m. Calculate the electric potential energy of the system.
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Four point charges, each with a magnitude of 1.00 C, are placed at the corners of a 45 rhombus with sides
of length 1.0 m. Calculate the electric potential energy of the system.
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- 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 = 210î V/m. Determine the following. (a) the change in the electric potential energy (in J) of the particle-field system (b) the electric potential difference (in V) through which the particle moves VSS f6 ssf60 ss 60 8. Three point-like charges are placed at the following points on the x-y system coordinates (q₁ is fixed at x = q2 is fixed at y = +3.00 cm, and q3 is fixed at x = +1.00 cm. Find the electric potential energy of the charge 92. Let 9₁ = -2.30 μC, 92 = -2.60 μC, and 93 = +4.00 μC. J f60 ssf60 ssf60 ssf60 ssf60 ss * y(cm) ) ssf60 ssf60 ssf60 ssf60 sc00 sxf60 ssf60 s +3 +2 + +1 O +1 +-1 --2 3 fe +2 ►x(cm) 30 ssf60 - 160 ssf60 ssf60 ( st ssf60 ssf60 ssfeesf6060 ssfó0 ssf60 ssf6 ssfen ssf6 ssf60 ssf60 ssf60 ssf60 ssf60 ssi -1.00 cm, ( f60 ssf60 ssf60 ssf60 ssf60 0 ssf60 ssfA particle has a charge of +1.76 μC and moves from point A to point B, a distance of 0.201 m. The particle experiences a constant electric force, and its motion is along the line of action of the force. The difference between the particle's electric potential energy at A and B is EPEA - EPEB = +8.87E-4 J. Calculate the magnitude of the electric force that acts on the particle.
- A uniform electric field of magnitude 4.3×105 N/CN/C points in the positive x direction. Find the change in electric potential energy of a 9.5-μCμCcharge as it moves from the origin to the point (6.0 mm, 0). Find the change in electric potential energy of a 9.5-μCμCcharge as it moves from the origin to the point (6.0 mm, 6.0 mm).A proton (mass = 1.67 x 10-27 kg, charge = 1.60 x 10-19 C) moves from point A to point B under the influence of an electrostatic force only. At point A the proton moves with a speed of 50 km/s. At point B the speed of the proton is 97.4 km/s. Determine the potential difference Vg-VA (in units of volts).Three equal point charges, each with charge 1.00 μCμC , are placed at the vertices of an equilateral triangle whose sides are of length 0.400 mm . What is the electric potential energy UUU of the system? (Take as zero the potential energy of the three charges when they are infinitely far apart.) Use ϵ0= 8.85×10−12 C^2/N⋅m^2 for the permittivity of free space.
- A particle with a charge of +4.20 nC is in a uniform elec tric field Ē directed to the left. It is released from rest and moves to the left: after it has moved 6.00 cm, its kinetic energy is found to be +1.50 x 10 6 J. (a) What work was done by the electric force? (b) What is the potential of the starting point with respect to the end point? (c) What is the magnitude of Ē?There are three point charges in a system. The first is -9.34 micro coulombs and is located at (0,51.5)m. The second is -9.45 micro coulombs and is located at (-97.4,0) m. The third is 5.65 micro coulombs and is located at (-20.4,52.9) m. What is the potential (in V) at the origin due to these three charges?Consider an electric field given by E = (4.0i + 3.0j)N/C.Determine the absolute value of the potential difference between the points: (x = 2.0m, y = 3.0m) and (x = 5.0m, y = 7.0m). 33.0 V 11.0 V 27.0 V 30.0 V 24.0 V
- An electrically charged particle is held at rest at the point x = 0; a second particle with equal charge is free to move along the positive x-axis. The potential energy of the system is U1x2 = C/x, where C is a positive constant that depends on the magnitude of the charges. Derive an expression for the x-component of force acting on the movable particle as a function of its position.A particle with a charge of q = 11.0 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 = 250î V/m. Determine the following. (a) the change in the electric potential energy (in J) of the particle-field system (b) the electric potential difference (in V) through which the particle moves VAn electron moving parallel to the x axis has an initial speed of 3.14 x 10° m/s at the origin. Its speed is reduced to 1.44 x 10 m/s at the point x = 2.00 cm. (a) Calculate the electric potential difference between the origin and that point. Volts (b) Which point is at the higher potential? O the origin O the point x 2.00 cm O both have the same potential Need Help? Read It Master It