Equipotential 5.033 V +1 nC -1 nC Sensors Electric Field Direction only Voltage ✔Values ✔Grid O At any point, the electric field is perpendicular to the equipotential line at that point, and it is directed toward lines of higher voltages. O At any point, the electric field is parallel to the equipotential line at that point. O At any point, the electric field is perpendicular o the equipotential line at that point, and it is directed toward lines of lower voltages. Place several E-Field Sensors at a few points on different equipotential lines, and look at the relationship between the electric t I and the equipotential lines. Which statement is true?
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- NoneThe potential difference across the capacitor C₂ in the adjoining figure is (C₁= 20 e.s.u., C₂ = 30 e.s.u., C3 =15 e.s.u. and V = 90 e.s.u.) (a) 10 e.s.u. (b) 30 e.s.u. (c) 40 e.s.u. (d) 20 e.s.u. C₁ C₂ C3 0←V→→→→0 Fig. Q. 7→A Click Submit to complete this assessment. Question 6 2.0µF and a 1.0µF capacitor are connected in parallel and a potential difference, AV , is applied across the combination. The 2.0 µF capacitor has: 2.0uF HI 1.OuF AV twice the potential difference of the 1.0µF capacitor. half the charge of the 1.0µF capacitor half the potential difference of the 1.0 uF capacitor a. Oc. O d. Twice the charge of the 1.0µF capacitor Click Submit to complete this assessment.
- A coaxial cable used in a transmission line has an inner radius of 0.076 mm and an outer radius of 0.89 mm. Calculate the capacitance per meter for the cable. ASsume that the space between the conductors is filled with a material with a dielectric constant of 2.8. Number i UnitsIn the figure given below calculate the potential difference between a and c by considering a path that contains R, ri, and E1. Given that E = 4.4V, E2 = 2.1V, r1= 2.3 N ,r2= 1.8 N and R= 5.5 2. Q4. Battery 1 Battery 2For a zero initial velocity, which of the following is true regarding the relationship of the final velocity to the potential difference used to accelerate the charge? O The potential difference is inversely proportional to the final velocity. O The potential difference is directly proportional to the square of the final velocity. O The potential difference is directly proportional to the final velocity. O The potential difference is inversely proportional to the square of the final velocity.
- B) Two voltage vectors Va and Vb combine to form a resultant vector Vr. Va = 110V Vb = 230 V Vb is at an angle of 45 degrees from the horizontal. Apply the cosine rule to determine the length of Vr and the angle between Va and Vr. Your Answer: Vr Va VbThe drawing shows a set of equipotential surfaces in cross section. The grid lines are 2.0 cm apart. Determine the electric field at position D, including direction. Take an upward-pointing electric field as positive, and a downward-pointing field as negative.In the figure below, equipotential lines are shown at 1-m intervals. Suppose that V₁ = 7 V₁ V₂ = 14 V, V3 = 21 V, V4 = 28 V, and V5 = 35 V. OV V₁ V₂ A V3 I Submit B V₁ 0 m 1 m 2m 3m 4 m 5 m Submit V₂ 1) What is the electric field at point A? Enter a positive value if the electric field points in the +x direction and a negative value if if the electric field points in the -x direction. (Express your answer to two significant figures.) V/m 2) What is the electric field at point B? Enter a positive value if the electric field points in the +x direction and a negative value if if the electric field points in the -x direction. (Express your answer to two significant figures.) V/m
- QUESTION A student forms a capacitor from two identical rectangular aluminum plates. The plates have dimensions 20 cm x 30 cm, and are spaced 1.0 mm apart. The student connects the plates to a 15-V battery. How much charge accumulates on one of the plates? ANSWER 34 nC 2.5 nC 0.75 DC 8.0 nC 0.50 nCFind the maximum potential difference between two parallel conducting plates separated by 0.5 cm of air, given the maximum sustainable electric field strength in air to be 3.0 x 106 V/m. Hint The maximum potential difference between them is Question Help: Message instructor Submit Question V.A -35.0 equipotential curve is located 4.5 cm from a 590.0 equipotential curve at the point where the curves are closest. What is the average strength of the electric field between the two equipotential curves around this point? Remember that 100. cm = 1.00 m. Use standard MKS unit abbreviations. Your Answer: Answer units