Physics for Scientists and Engineers
6th Edition
ISBN: 9781429281843
Author: Tipler
Publisher: MAC HIGHER
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Chapter 26, Problem 80P
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The ratios
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A proton moves at 5.20 x 105 m/s in the horizontal direction. It enters a uniform vertical electric field with a magnitude of 8.40 × 10³ N/C. Ignore any gravitational effects.
(a) Find the time interval required for the proton to travel 6.00 cm horizontally.
1.15e-7
☑
Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. ns
(b) Find its vertical displacement during the time interval in which it travels 6.00 cm horizontally. (Indicate direction with the sign of your answer.)
5.33e-3
☑
Your response is off by a multiple of ten. mm
(c) Find the horizontal and vertical components of its velocity after it has traveled 6.00 cm horizontally.
| ↑ +
jkm/s
A proton moves at 5.20 105 m/s in the horizontal direction. It enters a uniform vertical electric field with a magnitude of 8.40 103 N/C. Ignore any gravitational effects.
(a) Find the time interval required for the proton to travel 6.00 cm horizontally.
(b) Find its vertical displacement during the time interval in which it travels 6.00 cm horizontally. (Indicate direction with the sign of your answer.)
The figure below shows the electric field lines for two charged particles separated by a small distance.
92
91
(a) Determine the ratio 91/92.
1/3
×
This is the correct magnitude for the ratio.
(b) What are the signs of q₁ and 92?
91 positive
92 negative
×
Chapter 26 Solutions
Physics for Scientists and Engineers
Ch. 26 - Prob. 1PCh. 26 - Prob. 2PCh. 26 - Prob. 3PCh. 26 - Prob. 4PCh. 26 - Prob. 5PCh. 26 - Prob. 6PCh. 26 - Prob. 7PCh. 26 - Prob. 8PCh. 26 - Prob. 9PCh. 26 - Prob. 10P
Ch. 26 - Prob. 11PCh. 26 - Prob. 12PCh. 26 - Prob. 13PCh. 26 - Prob. 14PCh. 26 - Prob. 15PCh. 26 - Prob. 16PCh. 26 - Prob. 17PCh. 26 - Prob. 18PCh. 26 - Prob. 19PCh. 26 - Prob. 20PCh. 26 - Prob. 21PCh. 26 - Prob. 22PCh. 26 - Prob. 23PCh. 26 - Prob. 24PCh. 26 - Prob. 25PCh. 26 - Prob. 26PCh. 26 - Prob. 27PCh. 26 - Prob. 28PCh. 26 - Prob. 29PCh. 26 - Prob. 30PCh. 26 - Prob. 31PCh. 26 - Prob. 32PCh. 26 - Prob. 33PCh. 26 - Prob. 34PCh. 26 - Prob. 35PCh. 26 - Prob. 36PCh. 26 - Prob. 37PCh. 26 - Prob. 38PCh. 26 - Prob. 39PCh. 26 - Prob. 40PCh. 26 - Prob. 41PCh. 26 - Prob. 42PCh. 26 - Prob. 43PCh. 26 - Prob. 44PCh. 26 - Prob. 45PCh. 26 - Prob. 46PCh. 26 - Prob. 47PCh. 26 - Prob. 48PCh. 26 - Prob. 49PCh. 26 - Prob. 50PCh. 26 - Prob. 51PCh. 26 - Prob. 52PCh. 26 - Prob. 53PCh. 26 - Prob. 54PCh. 26 - Prob. 55PCh. 26 - Prob. 56PCh. 26 - Prob. 57PCh. 26 - Prob. 58PCh. 26 - Prob. 59PCh. 26 - Prob. 60PCh. 26 - Prob. 61PCh. 26 - Prob. 62PCh. 26 - Prob. 63PCh. 26 - Prob. 64PCh. 26 - Prob. 65PCh. 26 - Prob. 66PCh. 26 - Prob. 67PCh. 26 - Prob. 68PCh. 26 - Prob. 69PCh. 26 - Prob. 70PCh. 26 - Prob. 71PCh. 26 - Prob. 72PCh. 26 - Prob. 73PCh. 26 - Prob. 74PCh. 26 - Prob. 75PCh. 26 - Prob. 76PCh. 26 - Prob. 77PCh. 26 - Prob. 78PCh. 26 - Prob. 79PCh. 26 - Prob. 80PCh. 26 - Prob. 81P
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- Please help me solve this one more detail, thanksarrow_forwardA dielectric-filled parallel-plate capacitor has plate area A = 20.0 ccm2 , plate separaton d = 10.0 mm and dielectric constant k = 4.00. The capacitor is connected to a battery that creates a constant voltage V = 12.5 V . Throughout the problem, use ϵ0 = 8.85×10−12 C2/N⋅m2 . Find the energy U1 of the dielectric-filled capacitor. The dielectric plate is now slowly pulled out of the capacitor, which remains connected to the battery. Find the energy U2 of the capacitor at the moment when the capacitor is half-filled with the dielectric. The capacitor is now disconnected from the battery, and the dielectric plate is slowly removed the rest of the way out of the capacitor. Find the new energy of the capacitor, U3. In the process of removing the remaining portion of the dielectric from the disconnected capacitor, how much work W is done by the external agent acting on the dielectric?arrow_forwardIn (Figure 1) C1 = 6.00 μF, C2 = 6.00 μF, C3 = 12.0 μF, and C4 = 3.00 μF. The capacitor network is connected to an applied potential difference Vab. After the charges on the capacitors have reached their final values, the voltage across C3 is 40.0 V. What is the voltage across C4? What is the voltage Vab applied to the network? Please explain everything in steps.arrow_forward
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