A) 01 is positive, 02 is negative, 03 is positive. B) 01 is negative, 02 is posi

Question

A) 01 is positive, 02 is negative, 03 is positive. B) 01 is negative, 02 is positive, 03 is negative. C) O1 is positive, 02 is positive, 23 is negative. D) All three charges are negative. E) All three charges are positive. 9. A piece of plastic has a net charge of +2.00 uC. How many more protons than electrons does this pice of plastic have? (e-1.60 10-19c A) 1.25 x 1013 B) 1.25 x 1019 C)2.50x 1013 D) 2.50× 1019 10. A 1.0-C point charge is 15 m from a second point charge, and the 10 electric force on one of them due to the other is I.0 N. What is the magnitude of the second charge? (k-14mo = 8.99 × 109 N-m2/C2) A) 25 C B) 1.0 C C) 10 nC D) 0.025 C E) 25 nC 11. An atomic nucleus has a charge of +40e. What is the magnitude of the 11 electric field at a distance of 1.0 m from the center of the nucleus? (k = 1 /4xe0 = 8.998 109 N-m2/C2, e = 1.60 x 10-19 C) A) 5.4 x 10-8 N/C B) 5.6 x 10-8 N/C C) 5.8 x 10-8 N/C D) 6.0 x 10-8 N/C E) 6.2 x 10-8 N/C 12. If the electric flux through a closed surface is zero, 12_ the electric field at points on that surface must be zero.

Explanation / Answer

8

we know that the field lines for positive charge are radially outward and for negative charge is radially inward so

or we can say the positive charges are sources of field lines and negative charges are sinks of field lines

From the given picture

q1 is positive charge , q2 is negative charge and q3 is positive charge

so the answer is option A

9.

given piece of plastic has the charge is Q = 2*10^-6 C

and we know that the charge is positive charge

means the plastic will have both +ve and -ve charges but the +ve charges are more in number

the fundamental charge is q = 1.6*10^-19 C

so the charge Q = N*q

N = Q/q

N = (2*10^-6)/(1.6*10^-19)

N = 1.25*10^13

so the answer is plastic containing 1.25*10^13 number of protons more than the number of electrons

10.

q1 = 1.0 c, r = 15 m, q2 = ?

F = 1.0 N

from Coulomb's law

F = kq1*q2/r^2

1 = 8.99*10^9*1*q2/15^2

solving for q2

q2 = 2.50*10^-8 C

q2 = 25*10^-9 C

q2 = 25 nC

so the answer is option E

11.

atomic nucleus charge q = +40*e = 40*1.6*10^-19 C

r = 1.0 m

electric field is E = kq/r^2

E = 8.99*10^9*40*1.6*10^-19 /(1^2) N/C

E = 5.7536*10^-8 N/C

E = 5.8*10^-8 N/C

so the answer is option C

12.

By definition of electric flux

electric flux phi_E = E*A cos theta =0

so E is also zero because A is non zero

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