標題:

AL physics ~ Electrostatics

發問:

please refer to the picturehttp://ihs.meric.hk/rforum.php/307915.jpgIn the figure, O is the centre of a circle of radius 3 m.COA and BOD are diameters which are at right angle to each other.Charges of 6 nC, -2 nC, 3 nC and 4 nC are placed at A, B, C, D respectively.Find the electric field intensity... 顯示更多 please refer to the picture http://ihs.meric.hk/rforum.php/307915.jpg In the figure, O is the centre of a circle of radius 3 m. COA and BOD are diameters which are at right angle to each other. Charges of 6 nC, -2 nC, 3 nC and 4 nC are placed at A, B, C, D respectively. Find the electric field intensity at O. ANS: The electric field intensity at O is 6.7 N/C at an angle of 63.4 (degree) with the x-axis. thanks for help.

最佳解答:

The E-field intensity along the diameter BOD is (6 x 10-9)/(4πε0 x 32) = (10-9)/(6πε0) from O to B. The E-field intensity along the diameter COA is (3 x 10-9)/(4πε0 x 32) = (10-9)/(12πε0) from O to C. Hence, by Pyth. thm, the resultant E-field intensity megnitude is: √{[(10-9)/(6πε0)]2 + [(10-9)/(12πε0)]2} = 6.7 N/C With direction: tan θ = [(10-9)/(6πε0)]/[(10-9)/(12πε0)] = 2 θ = 63.4 degree

其他解答:

aa.jpg

 

此文章來自奇摩知識+如有不便請留言告知

arrow
arrow
    文章標籤
    名字 更多 插曲 文章 甜心
    全站熱搜

    hzb53jl55v 發表在 痞客邦 留言(0) 人氣()

    E-mail轉寄