A parallel plate capacitor has a surface charge density of magnitude o = 0.00433 C/m² on each plate.A small sphere, charged to Q = 9.40 × 10-6 C and located between the plates, undergoes a displacement of magnituded = 0.143 m. Refer to the figure, which is not drawn to scale.y+++++++Using the symbols in the palette, not their numeric equivalents, enter an expression, in Cartesian unit-vector notation, for the electric field within the capacitor.If the displacement is from left to right, then how much work, in joules, is performed on the sphere by the electric field?At what speed, in meters per second, is the sphere moving when it completes the displacement if it starts from rest and its mass is 0.115 kg?X

To solve this problem, we need to consider the relationship between work, the electric field, and…

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Example: A proton is released from rest at the positive plate of a capacitor. What is the proton's speed when it reaches the negative plate? The capacitor is illustrated below with spacing d=2.1mm. The strength of the electric field is E=4.5x104N/C. E proton + d=2.1mm Q [6.26] First off, note that the acceleration x component is negative because it points to the left so ax=-qE/m. We choose the velocity squared equation above because we want to solve for the final velocity (and we are not given the time interval): 2Eq(x-x₁) 2Eqxf : v²i + 2ax (xƒ − x₁) = 0 - - m m Where we made x-0 such that xF-0.0021m and that the proton starts at rest so vxF0. We have to take the square root to get the final speed: 2Eqri Vrf = ± √ m And substituting the numbers we get 4.5 x 10¹N -2( C -)(1.602x10-19C) (-0.0021m) 1.67x10-27kg Vxf = ± = ±1.35 x 105 m/s Notice that the sign underneath the radical is positive (as required) and to get the correct velocity component we need to choose the negative sign…
A parallel plate capacitor has the space between its plates filled with 1.4 mm of paper and 3.9 mm of teflon, as shown in the figure on the right. If the area of the plate of a capacitor is 9.2 cm², what is the capacitance of the capacitor? (Express your answer in pF (i.e. picofarads) Answer:
The plates of a parallel-plate capacitor are 3.22 mm apart, and each has an area of 9.92 cm². Each plate carries a charge of magnitude 4.50x10-8 C. The plates are in vacuum. ▼ What is the capacitance? Express your answer with the appropriate units. C = Submit Part B PHÅ V = Value Request Answer What is the potential difference between the plates? Express your answer with the appropriate units. 98 Value C μᾶ Units Units ? ?
The capacitor is charged to a field of 1.50 × 104 N/C and has plate separation of 5.00 cm. The plates each have length 1.00 m. An electron and a proton are each projected upward from the position shown with speed 3.00 × 106 m/s. a) Sketch the path followed by each charge. b) Where does each charge strike the capacitor? c) What is the speed of each charge when it strikes?
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A shark is able to detect the presence of electric fields as small as 1.70 μV/m. To get an idea of the magnitude of this field, suppose you have a parallel plate capacitor connected to a 2.00-V battery. How far apart must the parallel plates be to have an electric field of 1.70 μV/m between the plates? need answer in km
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