explain the motion of charged particles in a uniform electric field at x and y coordinates (two dimensions)
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explain the motion of charged particles in a uniform electric field at x and y coordinates (two dimensions)
Basic Details
The initial velocity of the particle along the x axis is v and along the y axis is zero. The acceleration along the axis can be determined by the force due to electric field. The motion of the particle along the axis can be determined by the second equation of motion.
Calculations
Step by step
Solved in 3 steps with 3 images
- A 4.1 MC point charge is placed on the x-axis at the point X₁ = -22.5m. A second, 2.7-MC point Charge is placed on the x-axis at the point X₂ = 10.3m. What is the x-component of the net electric field (in N/C) at the origin? (Remember that vector components can be positive or negative) N/C= A particle with charge 9₁ = +4.7 μC is located at (x=0,y=0). A second particle with charge 92 is located at (x=0,y=4.00 cm), and a third charge 93 = +5.4 µC is located at (x=3.00 cm,y=0). (a) In your notebook, draw a diagram of the three-charge system showing the location of the charges. (b) Calculate the potential energy of this three-charge system. PE = J -8 μCA point charge located at the origin generates an electric field that has the value E = (830 N/C) i + (150 N/C) j at the point (x, y) = (2.50 m, 3.20 m). What is the value of the charge?
- The figure below shows a charged particle, with a charge of q = +38.0 nC, that moves a distance of d = 0.185 m from point A to point B in the presence of a uniform electric field E of magnitude 245 N/C, pointing right. A positive point charge q is initially at point A, then moves a distance d to the right to point B. Electric field vector E points to the right. (a) What is the magnitude (in N) and direction of the electric force on the particle? magnitude Ndirection ---Select--- toward the right toward the left The magnitude is zero. (b) What is the work (in J) done on the particle by the electric force as it moves from A to B? J (c) What is the change of the electric potential energy (in J) as the particle moves from A to B? (The system consists of the particle and all its surroundings.) PEB − PEA = J (d) What is the potential difference (in V) between A and B? VB − VA = VA charged plastic ball of mass 5.00g is placed in a uniform electric field pointing vertically upwards with a strength of 300.0NC. Calculate the magnitude and sign of the charge required on the ball in order to create a force upwards that exactly equals the weight force of the ball.An electric dipole consists of two charges, +4 μC and -4 μC, separated by a distance of 2 cm. Determine the electric field strength at a point on the axial line of the dipole, 4 cm away from its center.
- A positively charged particle of mass 50 grams and charge 10 µC is released from rest at the origin in the uniform electric field of 100 N/C directed along the +X-axis. Determine its speed at x = 10 cm position.What must the charge (sign and magnitude) of a 2.45 g particle be for it to remain stationary when placed in a downward-directed electric field of magnitude 650 N/C ? Express your answer in microcoulombs.Two positive point charges, each 15 μC, lie along the x-axis at x = –0.15 m and x = +0.15 m. Find the electric field at (a) the origin (0,0) and (b) the point (0, 0.20 m) on the y-axis. Include a diagram.