如图一个匀强电场，电荷从a运动到b，为什么这段过程电场力做得功是a电势能减去b电势能，而不是b减去 如图所示,为一匀强电场,某带点粒子从A点运动到B点.在这一运...

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A\u3002\u8be5\u7c92\u5b50\u5e26\u6b63\u7535\uff0c\u56e0\u4e3a\u662f\u6cbf\u7535\u573a\u7ebf\u65b9\u5411\u8fd0\u52a8
B\u3002\u8be5\u7c92\u5b50\u5728A\u70b9\u7684\u7535\u52bf\u80fd\u6bd4\u5728B\u70b9\u7684\u7535\u52bf\u80fd\u591a1.5J\u3002\u56e0\u4e3a\u7269\u4f53\u6cbf\u7535\u573a\u7ebf\u65b9\u5411\u8fd0\u52a8\uff0c\u7535\u573a\u529b\u505a\u6b63\u529f\uff0c\u7535\u52bf\u80fd\u51cf\u5c0f\u3002
C\u3002\u7c92\u5b50\u5728A\u70b9\u7684\u52a8\u80fd\u6bd4\u5728B\u70b9\u7684\u52a8\u80fd\u5c111.5J\u3002\u56e0\u4e3a\u7535\u573a\u529b\u505a\u6b63\u529f\uff0c\u7535\u52bf\u80fd\u51cf\u5c0f\uff0c\u52a8\u80fd\u589e\u52a0\uff0c\u7535\u52bf\u80fd\u7684\u51cf\u5c0f\u91cf\u7b49\u4e8e\u52a8\u80fd\u7684\u589e\u52a0\u91cf\u3002
D\u3002\u6b63\u786e\u3002\u673a\u68b0\u80fd\u7684\u53d8\u5316\u91cf\u5c31\u662f\u52a8\u80fd\u7684\u53d8\u5316\u91cf\u3002\u52a8\u80fd\u7684\u589e\u52a0\u91cf\u7b49\u4e8e\u7535\u52bf\u80fd\u7684\u51cf\u5c0f\u91cf\u53731.5J\u3002

因为电势沿电场线的方向递减，所以由初剪末。而且电势是不可能有负数的，可以末剪初加绝对值呗

因为运动的话需要做功，而这些功正是a到b所降低的电势能提供的，所以是a-b

首先明确带什么电？在根据电场线方向判断做负功还是正的，就可以得出是a-b还是b-a
其实两种表达都可以，一把正确的表示是Wb-Wa，如果结果是正的，就可以知道做的是正功，反之亦然。

W_AB=qEd=qu_AB=q(φ_A-φ_B)= qφ_A- qφ_B=E_A- E_{B  所以是A 的电势能减去B的电势能}

初状态减末状态

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