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兩小球質量皆為m 且帶電量相等 各以長L的細線懸著 將其中一小球固定並使其細線成鉛直狀 此時兩線張角為Q 則另一小球(沒固定的)受靜電力F為何?

我畫完圖後設T為線的張力
F*(cos(Q/2)=T*sinQ
mg=T*cosQ
只算出F=mg*(tanQ/(cos(Q/2)))

而答案卻是F=2mg*(sin(Q/2))

請替我解答一下 這題到底怎麼算==
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