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- P1.30. Electric force on a test charge in the field of six point charges. Six point charges, each of value Q, are situated at (d, 0,0), (-d, 0, 0), (0, d, 0), (0, –d, 0), (0, 0, d), and (0,0, –d). A test charge q located at the origin is displaced by a distance A < d along the positive x-axis. Find an approximate expression for the electric force acting on the charge.5.1.A spherical conductor has a spherical cavity with the same center point.at the center of the cavity there is a point charge of q. If a total charge of Q is placed on the conductor ,what is Q i ,the total charge on the inner surface of the conductor? 2. Q o,the total charge on the outer surface of the conductor? 3.If the cavity has a radius of a and the conductor has a radius of b,what is the surface charge density on the inner surface of the conductor? 4.The outer surface of the conductor? 5.Assume that the charges on the inner and outer surface are uniformly distributed.Draw the electric field lines on the diagramQ. 1. Define electric flux and write its SI unit. The electric field components in the figure shown are: + CORDO a Ex X Z 100 N E₁ = ox, Ey = 0, E₂ = 0 where a = Cm the charge within the cube, assuming a = 0.1 m. · Calculate
- part 1 of 2 Consider the field due to a uniformly charged disk of radius R and charge Q. Along the symmetry axis at distance z from the cen- ter, the field has been written in following forms. where Eexact = Eo (1 。 (1-√R²+²) (1-7) E₁ ≈ Eo (1 R part 2 of 2 Eo = (Q/A) 280 To compare the accuracy of different ap- proximations it is convenient to work with the normalized difference to be specified below. The nth order smallness is charac- terized by e" and € = 3 with e being the smallness parameter. In general the nth or- der term takes the form Ce", where C is some finite constant (e.g 0.1Figure 1.52 shows a spherical shell of charge, of radius a and surface density σ, from which a small circular piece of radius b << a has been removed. What is the direction and magnitude of the field at the midpoint of the aperture? Solve this exercise in three ways: a) direct integration, b) by superposition, and c) using the relationship for a force on a small patch.In these experiment, you will be analyzing the parallel plate capacitor which is represented below. The area of the capacitor and the separation between the plates are adjustable. The capacitance of the setup can be calculated from: C = Kair€oh × L/d. In the setup, you will analyze dependance of C on Area, and on d. Area will be adjusted by changing the length L. HINT: Recall that K = 1.00059 ~ 1, and ɛo = 8.85 x 10-12 F/m (C²/Nm2). |CAUTION: You will be reading the capacitance value by using a multimeter. • Arrange the scale of the multimeter to the smallest possible scale. • Before each reading, always discharge the capacitor by connecting the plates with a wire.Select four random/arbitrary points (two near Q1 (left charge) and two near Q2 (right charge). Let Q1 = +50 microCoulombs (μ C) and Q2 = +50 microCoulombs (μ C). Calculate the Electric Field vectors at those points. Then overlay those Electric Field vectors on to the Electric Field Lines below (figure). Define your coordinate system such that the position of Q1 is at your origin with the +x axis horizontal to the right and the +y axis vertical pointing up. The position of Q2 is 52 cm to the right of your origin (Q1). Directly above the Q1 (along the +y axis), the maximum distance shown is 25 cm. Remember your E-field vectors should be tangent to the E-field lines with arrow lengths drawn proportionately.2.1. Four 10nC positive charges are located in the z = 0 plane at the corners of a square 8cm on a side. A fifth 10nC positive charge is located at a point 8cm distant from the other charges. Calculate the magnitude of the total force on this fifth charge for e = €0: Arrange the charges in the xy plane at locations (4,4), (4,-4), (-4.4), and (-4,-4). Then the fifth charge will be on the z axis at location z = 4/2, which puts it at 8cm distance from the other four. By symmetry, the force on the fifth charge will be z-directed, and will be four times the z component of force produced by each of the four other charges.A sphere of radius R has total charge Q. The volume charge density (C/m') within the sphere is p(r) = C/r² , where C is a constant to be determined. 1. Use the expression to find a magnitude of an electric field strenght E : a) inside the sphere (rConsider a uniformly charged ring in the xy plane, centered at the origin. The ring has radius a and positive charge q distributed evenly along its circumference. What is the direction of the electric field at any point on the z axis? What is the magnitude of the electric field along the positive z axis? Use k in your answer, where k=1/4πϵ0.Let q₁ the figure. Calculate the electric field at the points:(0; 0), (4; 0), (-3; 0) y (1; 0).Draw the vector E. == 3x10 ℃ and q₂ = 5x10 °C and q3 = -5x10 °C distributed as indicated in q1 (-3; 0) (0; 0) 93 (4; 0) 92 X (m)Consider a right triangle ABC with the right triangle at vertex B. The charges at A, at B, and at C, are known to be 5 mC, 4 mC, and 7 mC, respectively. Given that the side AB is numerically equal to the last two digits of your student number, in meters, and AC is thrice AB, find the magnitudes of the force and of the electric field at A. When AB = 12SEE MORE QUESTIONSRecommended textbooks for youCollege PhysicsPhysicsISBN:9781305952300Author:Raymond A. Serway, Chris VuillePublisher:Cengage LearningUniversity Physics (14th Edition)PhysicsISBN:9780133969290Author:Hugh D. Young, Roger A. 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