PLEASE HELP WITH AS MANY AS POSSIBLE... I AM TERRIBLY CONFUSED.
1. A charge moves a distance of 1.7 cm in the
direction of a uniform electric field having
a magnitude of 200 N/C. The electrical
potential energy of the charge decreases by
9.69095 × 10−19 J as it moves.
Find the magnitude of the charge on the
moving particle. (Hint: The electrical poten-
tial energy depends on the distance moved in
the direction of the field.)
Answer in units of C
2. Initially, both metal spheres are neutral.
In a charging process, 1 × 1013 electrons are
removed from one metal sphere and placed on
a second sphere. Then the electrical poten-
tial energy associated with the two spheres is
found to be −0.061 J .
The Coulomb constant is 8.98755 ×
109 N · m2/C2 and the charge on an electron
is 1.6 × 10−19 C.
What is the distance between the two
spheres?
Answer in units of m.
3. It takes 119 J of work to move 2.4 C of charge
from a positive plate to a negative plate.
What voltage difference exists between the
plates?
Answer in units of V
4The magnitude of a uniform electric field be-
tween the two plates is about 2 × 105 N/C.
If the distance between these plates is
0.1 cm, find the potential difference between
the plates.
Answer in units of V
5A force of 3.60 × 10−2 N is needed to move a
charge of 56.0 μC a distance of 25.0 cm in the
direction of a uniform electric field.
What is the potential difference that will
provide this force?
Answer in units of V.
6An electron moves from one plate of a capaci-
tor to another, through a potential difference
of 2495 V.
a) Find the speed with which the electron
strikes the positive plate.
Answer in units of m/s.
7b) If a proton moves from the positive plate to
the negative plate, find the speed with which
the proton strikes the negative plate.
Answer in units of m/s.
008
A proton is accelerated from rest through a
potential difference of 119 V.
Calculate the final speed of this proton.
Answer in units of m/s.
9. The three charges shown in the figure are
located at the vertices of an isosceles triangle.
The Coulomb constant is 8.98755 ×
109 N · m2/C2 and the acceleration of grav-
ity is 9.8 m/s2 .
4.1 cm
4.1 cm
1.4 cm
+
− − 3.3 × 10−9 C 3.3 × 10−9 C
5.1 × 10−9 C
Calculate the electric potential at the mid-
point of the base if the magnitude of the posi-
tive charge is 5.1×10−9 C and the magnitude
of the negative charges are 3.3 × 10−9 C.
Answer in units of V.
10. An electron that is initially 54 cm away from
a proton is displaced to another point.
The Coulomb constant is 8.98755 ×
109 Nm2/C2 and the acceleration of gravity
is 9.8 m/s2.
If the change in the electrical potential
energy as a result of this movement is
2.7 × 10−28 J, what is the final distance be-
tween the electron and the proton?
Answer in units of m.
11A potential difference of 117.0 V exists across
the plates of a capacitor when the charge on
each plate is 429.0 μC.
What is the capacitance?
Answer in units of F.
12A parallel-plate capacitor has a plate area of
188 cm2 and a plate separation of 0.0420 mm.
The permittivity of a vacuum is 8.85419 ×
10−12 C2/N · m2.
a) Determine the capacitance.
Answer in units of F.
13b) Determine the potential difference when
the charge on the capacitor is 530.0 pC.
Answer in units of V.
14A parallel-plate capacitor has a capacitance
of 0.28 μF and is to be operated at 6500 V.
a) Calculate the charge stored.
Answer in units of C
15b) What is the electrical potential energy
stored in the capacitor at the operating po-
tential difference?
Answer in units of J.
It is 2013 and this is easy for 8 th graders
1. To find the magnitude of the charge on the moving particle, you can use the formula for electrical potential energy: E = qV, where E is the electrical potential energy, q is the charge, and V is the potential difference. In this case, the potential difference is equal to the magnitude of the electric field multiplied by the distance moved: V = E/(F*d), where F is the magnitude of the electric field and d is the distance moved. Plugging in the given values, we have: V = (9.69095 × 10^-19 J)/((200 N/C)*(0.017 m)).
2. The distance between the two spheres can be found using the formula for electrical potential energy: E = k(q1*q2)/r, where E is the electrical potential energy, k is the Coulomb constant, q1 and q2 are the charges on the spheres, and r is the distance between them. Rearranging the formula, we have: r = k(q1*q2)/E. Plugging in the given values, we have: r = (8.98755 × 10^9 N·m^2/C^2)*(1.6 × 10^-19 C)^2/(-0.061 J).
3. The voltage difference between the plates can be found using the formula: V = W/q, where V is the voltage difference, W is the work done, and q is the charge. Plugging in the given values, we have: V = 119 J/2.4 C.
4. The potential difference between the plates can be found using the formula: V = Ed, where V is the potential difference, E is the magnitude of the electric field, and d is the distance between the plates. Plugging in the given values, we have: V = (2 × 10^5 N/C)*(0.001 m).
5. The potential difference can be found using the formula F = qV/d, where F is the force, q is the charge, V is the potential difference, and d is the distance. Rearranging the formula, we have: V = Fd/q. Plugging in the given values, we have: V = (3.60 × 10^-2 N)*(0.25 m)/(56.0 × 10^-6 C).
6a. The speed of the electron can be found using the formula: v = sqrt(2qV/m), where v is the speed, q is the charge, V is the potential difference, and m is the mass. Plugging in the given values, we have: v = sqrt((2)(1.6 × 10^-19 C)(2495 V)/(9.11 × 10^-31 kg).
7b. The speed of the proton can be found using the same formula as in 6a, but using the mass and charge of a proton. Plugging in the given values, we have: v = sqrt((2)(1.6 × 10^-19 C)(2495 V)/(1.67 × 10^-27 kg).
8. The final speed of the proton can be found using the formula: v = sqrt(2qV/m), where v is the speed, q is the charge, V is the potential difference, and m is the mass. Plugging in the given values, we have: v = sqrt((2)(1.6 × 10^-19 C)(119 V)/(1.67 × 10^-27 kg).
9. The electric potential at the midpoint of the base can be found using the formula: V = k*q/r, where V is the electric potential, k is the Coulomb constant, q is the charge, and r is the distance. Plugging in the given values, we have: V = (8.98755 × 10^9 N·m^2/C^2)*((5.1 × 10^-9 C - 3.3 × 10^-9 C)/0.041 m).
10. The final distance between the electron and the proton can be found using the formula for electrical potential energy: E = k(q1*q2)/r, where E is the electrical potential energy, k is the Coulomb constant, q1 and q2 are the charges, and r is the distance. Rearranging the formula, we have: r = k(q1*q2)/E. Plugging in the given values, we have: r = (8.98755 × 10^9 N·m^2/C^2)*((1.6 × 10^-19 C)^2 - (1.6 × 10^-19 C)(54 cm))/(2.7 × 10^-28 J).
11. The capacitance can be found using the formula: C = q/V, where C is the capacitance, q is the charge, and V is the potential difference. Plugging in the given values, we have: C = (429.0 × 10^-6 C)/(117.0 V).
12a. The capacitance can be found using the formula: C = ε0*(A/d), where C is the capacitance, ε0 is the permittivity of vacuum, A is the plate area, and d is the plate separation. Plugging in the given values, we have: C = (8.85419 × 10^-12 C^2/(N·m^2))*(0.0188 m^2)/(0.0420 × 10^-3 m).
13b. The potential difference can be found using the formula: V = q/C, where V is the potential difference, q is the charge, and C is the capacitance. Plugging in the given values, we have: V = (530.0 × 10^-12 C)/(0.0420 × 10^-6 F).
14a. The charge stored can be found using the formula: Q = CV, where Q is the charge stored, C is the capacitance, and V is the potential difference. Plugging in the given values, we have: Q = (0.28 × 10^-6 F)*(6500 V).
15b. The electrical potential energy stored in the capacitor can be found using the formula: E = (1/2)CV^2, where E is the electrical potential energy, C is the capacitance, and V is the potential difference. Plugging in the given values, we have: E = (1/2)*(0.28 × 10^-6 F)*(6500 V)^2.
1. To find the magnitude of the charge on the moving particle, we can use the equation for electrical potential energy: U = q * V, where U is the electrical potential energy, q is the charge, and V is the electrical potential difference.
In this case, we know the decrease in electrical potential energy (U) is 9.69095 × 10^(-19) J and the distance moved (d) is 1.7 cm. The electrical potential difference (V) can be calculated using the equation V = U / q.
First, convert the distance moved from centimeters to meters: d = 1.7 cm = 0.017 m.
Next, solve for V: V = U / q = (9.69095 × 10^(-19) J) / q.
Since the magnitude of the electric field is given as 200 N/C, we can also use the equation V = E * d, where E is the magnitude of the electric field and d is the distance moved in the direction of the field.
Solve for V using the equation V = E * d: V = (200 N/C) * (0.017 m).
Now we have two expressions for V, so we can set them equal to each other and solve for q: (9.69095 × 10^(-19) J) / q = (200 N/C) * (0.017 m).
Solve for q: q = (9.69095 × 10^(-19) J) / [(200 N/C) * (0.017 m)].
2. To find the distance between the two spheres, we can use the equation for electrical potential energy: U = k * (q1 * q2) / r12, where U is the electrical potential energy, k is the Coulomb constant, q1 and q2 are the charges on the spheres, and r12 is the distance between the centers of the spheres.
In this case, we know the electrical potential energy (U) is -0.061 J, the Coulomb constant (k) is 8.98755 × 10^9 N · m^2/C^2, and the charge on an electron (q) is 1.6 × 10^(-19) C.
Solve for r12 using the equation U = k * (q1 * q2) / r12: -0.061 J = (8.98755 × 10^9 N · m^2/C^2) * (1.6 × 10^(-19) C)^2 / r12.
Simplify the equation: -0.061 J = (8.98755 × 10^9 N · m^2/C^2) * (2.56 × 10^(-38) C^2) / r12.
Now we can solve for r12: r12 = (8.98755 × 10^9 N · m^2/C^2) * (2.56 × 10^(-38) C^2) / -0.061 J.
3. The voltage difference between the plates can be found using the equation V = U / q, where V is the voltage difference, U is the work done, and q is the charge moved.
In this case, we know the work done (U) is 119 J and the charge moved (q) is 2.4 C.
So, the voltage difference (V) is V = 119 J / 2.4 C.
4. To find the potential difference between the plates, we can use the equation V = E * d, where V is the potential difference, E is the magnitude of the electric field, and d is the distance between the plates.
In this case, we know the magnitude of the electric field (E) is 2 × 10^5 N/C and the distance between the plates (d) is 0.1 cm.
Convert the distance between the plates from centimeters to meters: d = 0.1 cm = 0.001 m.
Now we can solve for V: V = (2 × 10^5 N/C) * (0.001 m).