Find the strength of the electric field inside the cavity.
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Inside a sphere of radius R charged with constant volume density p, there is a spherical cavity of radius r, in which there are no charges. The center of the cavity is displaced relative to the center of the sphere by a distance a (a + r <R). Find the strength of the electric field inside the cavity.

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- The entire y axis is covered with a uniform linear charge density 4.9 nC/m. Determine the magnitude of the electric field on the x axis at x = 7.4 m.The entire y axis is covered with a uniform linear charge density 2.2 nC/m. Determine the magnitude of the electric field on the x axis at x= 5.5 m.Five charged particles are equally spaced around a semicircle of radius 100 mm, with one particle at each end of the semicircle and the remaining three spaced equally between the two ends. The semicircle lies in the region x<0 of an xy plane, such that the complete circle is centered on the origin. If each particle carries a charge of 6.00 nC , what is the electric field at the origin? Where could you put a single particle carrying a charge of -5.00 nC to make the electric field magnitude zero at the origin?
- Four charged particles are located at the corners of a square of side length a = 30.0 cm, as shown. The unit charge q =+10.0μC. (a) What is the magnitude and direction of the electric field at the lower left corner?The electric field at point P is 9.3 x 1020 N/C @ 343°. What is the angle of the force experienced by a proton if it is placed at point P. Express your answer in degrees.A square window of side length d lies in the y-z plane. The electric field everywhere in the area of the window is given by = ai + bj+ ck , where a = 4.00 N/C, b = 8.00 N/C, and c = 15.0 N/C. (a.) Which of the following area vectors describes the area and orientation of the window? (b.) Let the side length of the window be d = 20.0 cm = 0.200 m. What is the electric flux through the window, in Nm2/C?
- At what distance along the central axis of a uniformly charged plastic disk of radius R = 1.31 m is the magnitude of the electric field equal to 1/3 times the magnitude of the field at the center of the surface of the disk?Two test charges are located in the x–y plane. If q1=−4.550 nC and is located at x1=0.00 m, y1=0.9200 m, and the second test charge has magnitude of q2=4.200 nC and is located at x2=1.500 m, y2=0.450 m, calculate the x and y components, Ex and Ey, of the electric field E in component form at the origin, (0,0). The Coulomb force constant is 1/(4??0)=8.99×109 N·m2/ C2.