68.In Fig.45, the parts of the very thin filament bent into a "U" shape, have a uniform linear charge density of >=+50 RC/m. (k-9-10² Nm²/C², a-10 cm) 1) Find the magnitude of the electric field at point P. a) 10.23 N/C b) 12.23 N/C c) 14.23 N/C d) 16.23 N/C II) Find the direction of the electric field, with respect to the horizontal. a) 23° 34' b)24° 34' c)25° 34' d)26° 34' +2 a a P 1-2 e) 18.23 N/C e)27° 34'

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68.In Fig. 45, the parts of the very thin filament bent into a "U" shape, have a uniform linear
charge density of 2=+50 RC/m. (k=9-10²_N•m²/C², a=10 cm)
1) Find the magnitude of the electric field at point P.
a) 10.23 N/C b) 12.23 N/C
c) 14.23 N/C
d) 16.23 N/C
II) Find the direction of the electric field, with respect to the horizontal.
a) 23° 34'
b)24° 34'
c)25° 34'
d)26° 34'
+2
a
a
+2
+λ a
P
Fig.45
a
1-2
e) 18.23 N/C
e)27° 34'
Transcribed Image Text:68.In Fig. 45, the parts of the very thin filament bent into a "U" shape, have a uniform linear charge density of 2=+50 RC/m. (k=9-10²_N•m²/C², a=10 cm) 1) Find the magnitude of the electric field at point P. a) 10.23 N/C b) 12.23 N/C c) 14.23 N/C d) 16.23 N/C II) Find the direction of the electric field, with respect to the horizontal. a) 23° 34' b)24° 34' c)25° 34' d)26° 34' +2 a a +2 +λ a P Fig.45 a 1-2 e) 18.23 N/C e)27° 34'
Expert Solution
Step 1: Given

Physics homework question answer, step 1, image 1

text Charge density: end text
lambda equals 50 cross times 10 to the power of negative 12 end exponent space C divided by m

text Half length: end text
a equals 10 space c m space equals space 0.1 space m

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