Calculate the magnitude of electric field due to an electric dipole of dipole moment 3.56.10^-29 Cm. at a point 25.4nm away along the bisector axis
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Calculate the magnitude of electric field due to an electric dipole of dipole moment 3.56.10^-29 Cm. at a point 25.4nm away along the bisector axis.
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- Find the x and y components of the electric field produced by q1 and q2 in the figure shown below at point A and point B. (Take q1 = 1.88 µC and q2 = −1.16 µC.) Point A Ex = Ey = Point B Ex = Ey =The value of the electric field at a distance of 27.9 m from a point charge is 75.9 N/C and is directed radially in toward the charge. What is the charge? The Coulomb constant is 8.98755 x 10° N . m²/C². Answer in units of C.could you solve this
- Three charges are positioned at the corners of a parallelogram as shown below. YA 1.0 m 30° 2Q (a) If Q = 8.6 µC, what is the electric field (in N/C) at the unoccupied corner? 1.13e17 N/C X -3.0 m- H (b) What is the force (in N) on a 5.1 µC charge placed at this corner? 5.76e11 NAn 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.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.
- magnitude direction BµC 60.0⁰ A μC No 0.500 m i (a) Three point charges, A = 1.70 μC, B = 7.25 μC, and C = -4.60 µC, are located at the corners of an equilateral triangle as in the figure above. Find the magnitude and direction of the electric field at the position of the 1.70 μC charge. N/C ° below the +x-axis Cuc (b) How would the electric field at that point be affected if the charge there were doubled? The magnitude of the field would be halved. The field would be unchanged. The magnitude of the field would double. The magnitude of the field would quadruple. Would the magnitude of the electric force be affected? OYes(a) Determine the electric field strength at a point 1.00 cm to the left of the middle charge shown in the figure below. (Enter the magnitude of the electric field only.) Three charges lie along a horizontal line. A 6.00 µC positive charge is on the left. 3.00 cm to its right is a 1.50 µC positive charge. 2.00 cm to the right of the 1.50 µC charge is a −2.00 µC charge. N/C(b) If a charge of −2.10 µC is placed at this point, what are the magnitude and direction of the force on it? magnitude N direction18. 6.0 m and total charge Q = 120 µC distributed uniformly along the length 8. Figure 5 shows a thin rod of length L of the rod. By direct integration, find the magnitude of the electric field at the point P in the figure. (s) Integral Figure 5 Call on X-axis no pat L/3 2 m 120ML. L L/2 3n a