Respuesta :
Answer:
A. 30.7cm
B. [tex]1.7*10^{-10}C[/tex]
C. The electric field is directed away from the point of charge
Explanation:
A.
[tex]because, E=\frac{v}{d} \\\\d= \frac{4.98}{16.2}\\\\ d = 0.307m\\\\d = 30.7 cm[/tex]
B.
Considering Gauss's law
[tex]EA = \frac{Q}{e}\\\\ where, e = pertittivity. space= 8.85* 10^{-12} Fm^{-1} \\\\A = surface. area. with.radius 0.307m\\Q= eEA = (8.85*10^{-12})(16.2)(4\pi)(0.307)^{2}\\\\= 1.7*10^{-10}C[/tex]
C. The electric field directed away from the point of charge when the charge is positive.
(a) the distance is 0.307 m
(b) the magnitude of the charge is [tex]Q=1.7\times10^{-10}C[/tex]
(c) the electric field is directed away from the point charge
Coulomb's Law:
Let us consider the point charge Q, the electric field (E) and potential difference (V) due to this charge at a distance r is given by:
[tex]E=k\frac{Q}{r^2}\;\;and\\\\V=k\frac{Q}{r}[/tex]
where k is Coulomb's constant
[tex]E=\frac{V}{r} \\\\r=\frac{V}{E}[/tex]
(a) from the data provided in the question
[tex]r=\frac{4.98}{16.2}\;m\\\\r=0.307\;m[/tex]
(b) now the charge can ve calculated as follows:
[tex]V=k\frac{Q}{r}\\\\Q=\frac{Vr}{k}\\\\Q=\frac{4.98\times0.307}{9\times10^9}C\\\\Q=1.7\times10^{-10}C[/tex]
(c) Since the charge is positive the electric field is directed away from the point charge,
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