2. Two uniformly charged rods of length L are placed perpendicularly as shown above. The rod on the left is negatively charged while the rod on top is positive charged. Assuming both rods have a uniform charge distribution A, find the next electric field in vector format (E) at point P.
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- Charge q1 = 10.5 nC is located at the coordinate system origin, while charge q2 = -3.5 nC is located at (a, 0), where a = 0.85 m. The point P has coordinates (a, b), where b = 3.5 m. Part A: At the point P, find the x-component of the electric field Ex in units of N/C. Part B: At the point P, find the y-component of the electric field Ey in units of N/C.The figure below shows a small, charged bead, with a charge of q = +41.0 nC, that moves a distance of d = 0.174 m from point A to point B in the presence of a uniform electric field E of magnitude 255 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 bead? magnitude Ndirection (b) What is the work (in J) done on the bead 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 bead moves from A to B? (The system consists of the bead and all its surroundings.) PEB − PEA = J (d) What is the potential difference (in V) between A and B? VB − VA = VTwo identical thin rods of length L are each charged with a positive charge q that is uniformly distributed on each rod. The rods are arranged parallel to each other along the x direction and are separated by a distance 2d, as shown in the figure below. Determine the electric field at point P, which is located a distance d from the right endpoint of each rod. Hint: You will eventually need the following integral: น 1 S (a²+u²)3/2 du = − + C √a²+u²
- The drawing shows a positive point charge q1, a second point charge q2 that may be positive or negative, and a spot labeled P, all on the same straight line. The distance d between the two charges is the same as the distance between q1 and the spot P. With q2 present, the magnitude of the net electric field at P is twice what it is when q1 is present alone. Given that q1 = 0.50 μ C, determine q2.A straight, nonconducting plastic wire 7.00 cm long carries a charge density of 155 nC/m distributed uniformly along its length. It is lying on a horizontal tabletop. Part A Find the magnitude and direction of the electric field this wire produces at a point 4.00 cm directly above its midpoint. Express your answer in newtons per coulomb. IVE ΑΣΦ E = Submit Part B X Incorrect; Try Again; 4 attempts remaining electric field is directed upward O electric field is directed downward Submit Part C Previous Answers Request Answer E = Correct Previous Answers Submit www. If the wire is now bent into a circle lying flat on the table, find the magnitude and direction of the electric field it produces at a point 4.00 cm directly above its center. Express your answer in newtons per coulomb. 195| ΑΣΦ Request Answer ? www. N/C ? N/C ✓A positively charged particle Q1=+25 nC is held fixed at the origin. A second charged Q2 of mass m=8.5 ug is floating a distance d=25 cm above charge Q1. The net force on Q2 is equal to zero. You may assume this system is close to the surface of the Earth. b. Calculate the magnitude of q2 in units of nanocoulumbs.
- A thin, insulating rod of length Icarries a linear charge density^(x) that varies with distance according to (x)=Ax (in SI units). The location of the origin is shown in the figure. A point charge q is located a distance I from the end of the rod as shown. a. What are the SI units of the constant A? b. Find the force that the rod exerts on q.È= [dE = [k9 .. Start with... f... dxConsider three charges q1 = 6.3 µC, q2 = 0.8 µC, and q3 = -2.3 µC, arranged as shown below. (a) What is the electric field at a point 1.0 cm to the left of the middle charge? N/C to the right or left(b) What is the magnitude of the force on a 2.2 µC charge placed at this point? N
- The figure below shows three regions in space separated by two infinite sheets of charge with surface charge densities σ and -σ. Find the magnitude and direction of the electric field in each region in terms of σ and ε. Given σ = 5 x10-6 C/m², find the acceleration vector components of an electron placed in Region 2. Region 1 Region 2 Region 3 BoldThe figure below shows a small, charged bead, with a charge of q = +42.0 nC, that moves a distance of d = 0.189 m from point A to point B in the presence of a uniform electric field E of magnitude 270 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 bead? magnitude Ndirection (b) What is the work (in J) done on the bead 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 bead moves from A to B? (The system consists of the bead and all its surroundings.) PEB − PEA = J (d) What is the potential difference (in V) between A and B? VB − VA = V