6. Six point charges of equal magnitude Q are held at the corners of a hexagon with the signs of the charges as shown. Each side of the hexagon has length 'a', and 'P' is at the centre of the hexagon. a -Q a) Derive the formula for the electric field strength at point P due to ONE point charge. No directions are necessary for this question. [T2] +Q O b) Draw the arrows to represent the direction of the field at P due to point charge A and point charge B. [T1] c) If the magnitude of Q is 2.6 µC and the length 'a' is 0.12 m, determine the magnitude and the direction of the electric field strength at point P due to all six charges. [T5] +QO A B Q ОР a Q O+Q
6. Six point charges of equal magnitude Q are held at the corners of a hexagon with the signs of the charges as shown. Each side of the hexagon has length 'a', and 'P' is at the centre of the hexagon. a -Q a) Derive the formula for the electric field strength at point P due to ONE point charge. No directions are necessary for this question. [T2] +Q O b) Draw the arrows to represent the direction of the field at P due to point charge A and point charge B. [T1] c) If the magnitude of Q is 2.6 µC and the length 'a' is 0.12 m, determine the magnitude and the direction of the electric field strength at point P due to all six charges. [T5] +QO A B Q ОР a Q O+Q
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P2
Answer 6
![5. A charge of - 2q is placed at the origin and a second charge of +q is placed at x = 5.0 cm. Where can a third
charge +q be placed so that it experiences a force of zero? [A5]
6. Six point charges of equal magnitude Q are held at the corners of a hexagon with the signs of the charges as
shown. Each side of the hexagon has length 'a', and 'P' is at the centre of the hexagon.
a
-Q
a) Derive the formula for the electric field strength at point P due to ONE
point charge. No directions are necessary for this question. [T2]
+ Q
●
b) Draw the arrows to represent the direction of the field at P due to point
charge A and point charge B. [T1]
c) If the magnitude of Q is 2.6 µC and the length 'a' is 0.12 m, determine
the magnitude and the direction of the electric field strength at point P
due to all six charges. [T5]
a
8. An electron with a horizontal speed of 4.7 x 106 m/s passes through two horizontal plates, as shown below. The
magnitude of the electric field between the plates is 150 N/C. The plates are 8.5 cm long.
a. Calculate the vertical component of the electron's final velocity [T5]
b. Calculate the final velocity of the electron. [T3]
e
X
2
+QQ
A
B
O
ОР
O+Q](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffc8ca1f5-10b5-4e76-9ca1-9785bb9dcba6%2F5472cd9d-f76c-46ba-bcc4-a6f791f97f1b%2Fnfo9gou_processed.jpeg&w=3840&q=75)
Transcribed Image Text:5. A charge of - 2q is placed at the origin and a second charge of +q is placed at x = 5.0 cm. Where can a third
charge +q be placed so that it experiences a force of zero? [A5]
6. Six point charges of equal magnitude Q are held at the corners of a hexagon with the signs of the charges as
shown. Each side of the hexagon has length 'a', and 'P' is at the centre of the hexagon.
a
-Q
a) Derive the formula for the electric field strength at point P due to ONE
point charge. No directions are necessary for this question. [T2]
+ Q
●
b) Draw the arrows to represent the direction of the field at P due to point
charge A and point charge B. [T1]
c) If the magnitude of Q is 2.6 µC and the length 'a' is 0.12 m, determine
the magnitude and the direction of the electric field strength at point P
due to all six charges. [T5]
a
8. An electron with a horizontal speed of 4.7 x 106 m/s passes through two horizontal plates, as shown below. The
magnitude of the electric field between the plates is 150 N/C. The plates are 8.5 cm long.
a. Calculate the vertical component of the electron's final velocity [T5]
b. Calculate the final velocity of the electron. [T3]
e
X
2
+QQ
A
B
O
ОР
O+Q
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