Two 10-cm-diameter charged disks face each other, 25 cm apart. The left disk is charged to - 50 nC and the right disk is charged to + 50 nC .
What is the electric field E, both magnitude and direction, at the midpoint between the two disks?
What is the force F on a -2.3 nC charge placed at the midpoint?
Physics Question with Disks?
2018-09-15 8:50 pm
回答 (2)
2018-09-15 9:04 pm
Electric field
The strength or magnitude of the field at a given point
is defined as the force that would be exerted on a
positive test charge of 1 coulomb placed at that point;
the direction of the field is given by the direction of
that force.
E = F/Q = kQ/r² Newtons/coulomb
k = 8.99e9 Nm²/C²
r = 25 cm/2 = 0.25 m/2 = 0.125 m
from left charge, E = k(50nC)/(0.125²) Newtons/coulomb pointed left
right charge, E is the same
total E = 2(8.99e9)(50e-9)/(0.125²) Newtons/coulomb pointed left
total E = (8.99)(6400) Newtons/coulomb pointed left
force F on a -2.3 nC charge
F = 2k(50nC)/(0.125²) Newtons/coulomb x 2.3 nC
F = 2(8.99e9)(50e-9)(2.3e-9)/(0.125²) Newtons pointed right
F = (8.99)(230)(64) nN pointed right
The strength or magnitude of the field at a given point
is defined as the force that would be exerted on a
positive test charge of 1 coulomb placed at that point;
the direction of the field is given by the direction of
that force.
E = F/Q = kQ/r² Newtons/coulomb
k = 8.99e9 Nm²/C²
r = 25 cm/2 = 0.25 m/2 = 0.125 m
from left charge, E = k(50nC)/(0.125²) Newtons/coulomb pointed left
right charge, E is the same
total E = 2(8.99e9)(50e-9)/(0.125²) Newtons/coulomb pointed left
total E = (8.99)(6400) Newtons/coulomb pointed left
force F on a -2.3 nC charge
F = 2k(50nC)/(0.125²) Newtons/coulomb x 2.3 nC
F = 2(8.99e9)(50e-9)(2.3e-9)/(0.125²) Newtons pointed right
F = (8.99)(230)(64) nN pointed right
2018-09-15 8:54 pm
You’re taking the class not us brah
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