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PIT_PIT [208]
3 years ago
15

What is the electric field (in N/C) at a point 5.0 cm from the negative charge and along the line between the two charges?

Chemistry
1 answer:
Natali [406]3 years ago
4 0

Answer: E = 2.455 x 10^5 N/C

Explanation:

q1 = 1.2x10^-7C

q2 = 6.2x10^-8C

Electric field, E = kQ/r²

where k = 9.0x10^9

since the location is (27 - 5)cm from q1

hence electric field, E1 = k*q1/r²

E1= (9x10^9 x 1.2x10^-7)/(0.22)² = 22314.05 N/C

for q2:

E1 = k*q2/r²

E2 at 5cm

E2 = (9x10^9 x 6.2x10^-8)/(0.05)² = 223200 N/C

Hence, the total electric field at 5cm position is

E = E1 + E2

E = 22314.05 + 223200 = 245514.05 N/C

E = 2.455 x 10^5 N/C

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In the following reaction, how many liters of o2 will produce 43.62 liters of co2 at stp? c3h8 5 o2 3 co2 4 h2o
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3 years ago
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Hope this helps ^-^

Explanation:

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A 30.0 L sample of nitrogen inside a rigid, metal container at 20.0 °C is placed inside an oven whose temperature is 50.0 °C. Th
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Answer:

3.31 atm.

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