A uniform electrical field is established by connecting the plates of a parallel

Question

A uniform electrical field is established by connecting the plates of a parallel-plate capacitor to a 11V battery. The plates of the capacitor are separated a distance 3.50 cm. What is the magnitude of the electrical field in the capacitor ? A positive charge of 6.0 x10^-6 C moves from the positive to the negative plate. What is the change in electrical potential energy for this charge Points A and B in the figure above are 0.80 cm apart. What is the electrical potential between point A and B ? Given that the shared area of the parallel plates is 4500 cm^2. What is the capacitance of this parallel plate capacitor ?

Explanation / Answer

) F = k*q1*q2/r^2 = 9.0x10^9*6.0x10^-6*3.0x10^-6/0.30^2 = 1.8N In electrostatistics, a positve sign of force indicates (a) repulsion between charges 3) An object continues to move in a straight line without a constant speed if no net force acts on it. This statement is: (d) none of the above 4) In what Direction does the frictional force act on an object as it slides down an inclined plane? (c) Parallel to the inclined surface (direction up the ramp) 5) What would be the gravitational force between the moon and the earth if the mass of the moon were 4 times greater, and its distance from the earth double the present value? (d) the same as the present value 6) when 2 resistances of equal value are connected in series, the net resistance (b) becomes double the original resistance 7_ the sum of all currents entering a circuit junction (d) equals the sum of the currents leaving the juction the below are not multiple choice - but anyone who helps me answer them will be voted best answer with 5 stars... Three charges are located along a single dimension. Charge A ( + 1.00 x 10 - 6 C) is located at the origin, x = 0m. Charge B (+2.00 x 10 -6 C) is located at x = .500m. Charge C (-2.50 x 10-6 C) is located at x=1.50 m a - find the total force (and direction) on charge B. F = k*qA*qB/rAB^2 + k*qB*qC/rBC^2 = 9.0x10^9*((1.00 x 10 - 6*2.00 x 10 -6)/(0.50^2) + (2.00 x 10 -6*2.50 x 10-6)/1^2) = 0.117N (to the right) b - What would be the force on charge B is Charge C was positive Now F = k*qA*qB/rAB^2 - k*qB*qC/rBC^2 = 9.0x10^9*((1.00 x 10 - 6*2.00 x 10 -6)/(0.50^2) - (2.00 x 10 -6*2.50 x 10-6)/1^2) = 0.027 N to the right An Oil drop weighs 1.9 x 10-14 N. It is suspended in an electric field of instensity 4.0 x 10 4 (little 4) N/C. a - what is the charge of the oil drop m*g = Eq So q = m*g/E = 1.9x10^-14/4.0x10^4 = 4.8x10^-19C b - If the drop is attracted toward a positive plate, how many excess electrons does it have? n = 4.8x10^-19/1.6x10^-19 = 3 excess electrons Two parallel plates are 0.500 m apart. the electric field between the plates is 6.00 x 10 3 (little 3) N/C. c - what is the potential difference between the plates? V = E*d = 6.00x10^3*0.500 = 3.00x10^3 V d - what work is done moving an electron from one plate to another in units of both joules an electron volts? U = V*q = 3.00x10^3*1.60x10^-19 = 4.80x10^-16J in eV = 4.80x10^-16J/1.60x10^-19 =3000eV Two charges, q1 and q2 (wierd symbols that look like q's) are seperated by a distance d and exert a force f on eachother. a - what new force will exist if i - q1 is doubled-------F is doubled ii - q1 and q2 are halved ----F is 1/4 iii - d is trippled?----F is 1/9 iv - q1 is tripled and d is doubled?---- F is 3/4

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